The minimum distance (absolute value) from the central maximum is approximately 8.55 × 10−5 mm.
(a)Fraction of maximum intensity at a distance of 600 cm from the central maximum of the interference. Consider that monochromatic light of wavelength λ is incident on and perpendicular to two slits separated by a distance d. This causes an interference pattern on a screen some distance away.
The pattern will have alternating light and dark fringes, with the central maximum being the brightest and the fringe intensities decreasing with distance from the central maximum.
The distance from the central maximum to the first minimum (the first dark fringe) is given by:$$sin\theta_1=\frac{\lambda}{d}$$$$\theta_1=\sin^{-1}\frac{\lambda}{d}$$Similarly, the distance from the central maximum to the nth minimum is given by:$$sin\theta_n=n\frac{\lambda}{d}$$$$\theta_n=\sin^{-1}(n\frac{\lambda}{d})$$At a distance x from the central maximum, the intensity of the interference pattern is given by:$$I(x)=4I_0\cos^2(\frac{\pi dx}{\lambda D})$$where I0 is the maximum intensity, D is the distance from the slits to the screen, and x is the distance from the central maximum. At a distance of 600 cm (or 6 m) from the central maximum, we have x = 6 m, λ = 656.3 nm = 6.563 × 10−7 m, d = 0.200 mm = 2 × 10−4 m, and we can assume that D ≈ 1 m (since the distance to the screen is much larger than the distance between the slits).
Substituting these values into the equation for intensity gives:$$I(6\ \text{m})=4I_0\cos^2(\frac{\pi (2\times 10^{-4})(6.563\times 10^{-7})}{(1)})$$$$I(6\ \text{m})=4I_0\cos^2(0.000412)$$$$I(6\ \text{m})=4I_0\times 0.999998$$$$I(6\ \text{m})\approx 4I_0$$Therefore, the intensity at a distance of 600 cm from the central maximum is approximately 4 times the maximum intensity.(b) Minimum distance (absolute in mm) from the central maximum where the intensity is at the value found in part (a)At the distance from the central maximum where the intensity is 4I0, we have x = 6 m and I(x) = 4I0.
Substituting these values into the equation for intensity gives:$$4I_0=4I_0\cos^2(\frac{\pi (2\times 10^{-4})(6.563\times 10^{-7})}{(1)})$$$$1=\cos^2(0.000412)$$$$\cos(0.000412)=\pm 0.999997$$$$\frac{\pi dx}{\lambda D}=0.000412$$$$d=\frac{0.000412\lambda D}{\pi x}$$$$d=\frac{0.000412(656.3\times 10^{-9})(1)}{\pi(6)}$$$$d\approx 8.55\times 10^{-8}$$The minimum distance from the central maximum where the intensity is 4 times the maximum intensity is approximately 8.55 × 10−8 m = 0.0855 μm = 8.55 × 10−5 mm.
Therefore, the minimum distance (absolute value) from the central maximum is approximately 8.55 × 10−5 mm.
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After a couple practice drops, do the first real drop and record the time in the space below. Then calculate and record the acceleration due to gravity. (You will have to do a kinematics problem.)
h = 2 m t₁ = 0.70 s t₂ = 0.58 s t3 = 0.62 s t4 = 0.73 s
t5 = 0.54 s
The acceleration due to gravity for this object is 6.8 m/s².
To calculate the acceleration due to gravity of an object, Using the kinematics and the formula below can be used; a = (2Δh) / t² Where; h = height, t = time, Δh = difference in height .
The time will be the average of the five attempts; (t₁+t₂+t₃+t₄+t₅)/5 = (0.7+0.58+0.62+0.73+0.54)/5 = 0.634 sΔh = 2m - 0m = 2ma = (2Δh) / t² = (2 * 2) / 0.634² = 6.8 m/s².
Kinematics is a discipline of physics and a division of classical mechanics that deals with the motion of a body or system of bodies that is geometrically conceivable without taking into account the forces at play (i.e., the causes and effects of the motions). The goal of kinematics is to offer a description of the spatial positions of bodies or systems of material particles, as well as the velocities and rates of acceleration of those velocities.
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(IN] w) p 20 19 18 17 16 15 14 13 12 11 10 3 -1 -2 0 1 1 2 3 4 AK The motion of a student in the hall 5 6 1. Describe the motion 2. Find the displacement in the north direction 3. Find the displacement in the south direction 4. Find the time it travelled north 7 t(s) 8 5. Find the time it travelled south 6. Find the total displacement 7. Find the total distance travelled 8. Find the total average velocity 9. Find the total average speed 10. At what instant did the object travelled the fastest? Explain. 11. At what time did the object travelled the slowest? Explain. 9 10 11 12 13
1. The motion of a student in the hall can be represented as follows: The student initially moves towards the north direction and reaches a maximum displacement of 5m. The student then turns back and moves towards the south direction and attains a maximum displacement of -2m.
The student then moves towards the north direction and attains a final displacement of 4m before coming to a stop.2. The displacement in the north direction can be calculated as follows:
Displacement = final position - initial position= 4 - 0 = 4mTherefore, the displacement in the north direction is 4m.
3. The displacement in the south direction can be calculated as follows: Displacement = final position - initial position= -2 - 5 = -7mTherefore, the displacement in the south direction is -7m.
4. The time it travelled north can be calculated as follows:
Time taken = final time - initial time= 8 - 0 = 8sTherefore, the time it travelled north is 8s.5.
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(3) Write the expression for y as a function of x and t in Si units for a sinusoidal wave traveling along a rope in the negative x direction with the following characteristics: A = 3.75 cm, 1 - 90.0 cm, f = 5.00 Hz, and yo, t) = 0 at t = 0. (Use the following as necessary: x and t.) v - 0.0875 sin (6.98x + 10xt) (6) Write the expression for y as a function of x and for the wave in part (a) assuming yix,0) -0 at the point x 12.5 cm (Use the following us necessary: x and ) y - 0.0875 sin (6.98x + 10x7 - 87.25) X
The expression for the wave function when y(x=12.5 cm, t) = 0;
y(x,t) = 3.75 sin (6.98x - 31.4t + π)
(a)The general expression for a sinusoidal wave is represented as;
y(x,t) = A sin (kx - ωt + φ),
where;
A is the amplitude;
k is the wave number (k = 2π/λ);
λ is the wavelength;
ω is the angular frequency (ω = 2πf);
f is the frequency;φ is the phase constant;
andx and t are the position and time variables, respectively.Now, given;
A = 3.75 cm (Amplitude)
f = 5.00 Hz (Frequency)y(0,t) = 0 when t = 0.;
So, using the above formula and the given values, we get;
y(x,t) = 3.75 sin (6.98x - 31.4t)----(1)
This is the required expression for the wave function in Si unit, travelling along the negative direction of x-axis.
(b)From part (a), the required expression for the wave function is;
y(x,t) = 3.75 sin (6.98x - 31.4t) ----- (1)
Let the wave function be 0 when x = 12.5 cm.
Hence, substituting the values in equation (1), we have;
0 = 3.75 sin (6.98 × 12.5 - 31.4t);
⇒ sin (87.25 - 6.98x) = 0;
So, the above equation has solutions at any value of x that satisfies;
87.25 - 6.98x = nπ
where n is any integer. The smallest value of x that satisfies this equation occurs when n = 0;x = 12.5 cm
Therefore, the expression for the wave function when y(x=12.5 cm, t) = 0;y(x,t) = 3.75 sin (6.98x - 31.4t + π)----- (2)
This is the required expression for the wave function in Si unit, when y(x=12.5 cm, t) = 0, travelling along the negative direction of x-axis.
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2 (a) A scientist measures the internal energy U in a gas as a function of temperature T. The quantities are found to be related by the equation 5A U = KBT0.5 + f(P,V), (1) 2 where A is a constant, and f(P, V) is a function of pressure and volume only. (i) Is this an ideal gas? Justify your answer in one or two sentences. (ii) What is the specific heat capacity of the gas for a constant volume process, cy? [Hint How did we calculate heat capacity cy for the ideal gas?] [3] [4]
The gas described by the equation is not an ideal gas because the relationship between internal energy U and temperature T does not follow the ideal gas law, which states that U is directly proportional to T.
(i) An ideal gas is characterized by the ideal gas law, which states that the internal energy U of an ideal gas is directly proportional to its temperature T. However, in the given equation, the internal energy U is related to temperature T through an additional term, f(P,V), which depends on pressure and volume. This indicates that the gas deviates from the behavior of an ideal gas since its internal energy is influenced by factors other than temperature alone.
(ii) The specific heat capacity at constant volume, cy, refers to the amount of heat required to raise the temperature of a gas by 1 degree Celsius at constant volume. The equation provided, 5A U = KBT^0.5 + f(P,V), relates the internal energy U to temperature T but does not directly provide information about the specific heat capacity at constant volume. To determine cy, additional information about the behavior of the gas under constant volume conditions or a separate equation relating heat capacity to pressure and volume would be required.
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Given an object distance of 12 cm and a lens with focal length
of magnitude 4 cm, what is the image distance for a concave lens?
Give your answers in cm.
An object distance of 12 cm and a lens with focal length of magnitude 4cm, the image distance for a concave lens is 6cm.
To calculate the image distance for a concave lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length of the concave lens (given as 4 cm)
v = image distance (unknown)
u = object distance (given as 12 cm)
Let's substitute the given values into the formula and solve for v:
1/4 = 1/v - 1/12
To simplify the equation, we can find a common denominator:
12/12 = (12 - v) / 12v
Now, cross-multiply:
12v = 12(12 - v)
12v = 144 - 12v
Add 12v to both sides:
12v + 12v = 144
24v = 144
Divide both sides by 24:
v = 6cm
Therefore, the image distance for a concave lens is 6cm.
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The following point charges are placed on the x axis: 2uC at x = 20cm; -3uC at x =30cm; -4 uC at x = 40 cm. Find
a) the total electric field at x=0
b) the total potential at x=0
c) if another 2uC charge is placed at x=0, find the net force on it
a) The electric field at x = 0 is given by the sum of the electric fields due to all the charges at x = 0.
The electric field due to each charge at x = 0 can be calculated as follows:
Electric field, E = Kq/r²
Here, K = Coulomb's constant = 9 × 10^9 Nm²/C², q = charge on the point charge in Coulombs,
r = distance between the point charge and the point where the electric field is to be calculated.
Distance between the first point charge (2 μC) and x = 0 = 20 cm = 0.2 m.
The electric field due to the first point charge at x = 0 is
E_1 = Kq1/r1²
= (9 × 10^9)(2 × 10^-6)/0.2²N/C
= 90 N/C
Distance between the second point charge (-3 μC) and x = 0 = 30 cm = 0.3 m.
The electric field due to the second point charge at x = 0 is
E_2 = Kq_2/r_2²
= (9 × 10^9)(-3 × 10^-6)/0.3²N/C
= -90 N/C
Distance between the third point charge (-4 μC) and x = 0 = 40 cm = 0.4 m.
The electric field due to the third point charge at x = 0 is
E_3 = Kq_3/r_3²
= (9 × 10^9)(-4 × 10^-6)/0.4²N/C
= -90 N/C.
The total electric field at x = 0 is the sum of E_1, E_2, and E_3.
E = E_1 + E_2 + E_3 = 90 - 90 - 90 = -90 N/C
Putting a negative sign indicates that the direction of the electric field is opposite to the direction of the x-axis.
Hence, the direction of the electric field at x = 0 is opposite to the direction of the x-axis.
b) Potential at a point due to a point charge q at a distance r from the point is given by:V = Kq/r.
Therefore, potential at x = 0 due to each point charge can be calculated as follows:
Potential due to the first point charge at x = 0 is
V_1 = Kq_1/r_1 = (9 × 10^9)(2 × 10^-6)/0.2 J
V_1 = 90 V
Potential due to a second point charge at x = 0 is
V_2 = Kq_2/r_2 = (9 × 10^9)(-3 × 10^-6)/0.3 J
V_2 = -90 V
Potential due to a third point charge at x = 0 is
V_3 = Kq_3/r_3
= (9 × 10^9)(-4 × 10^-6)/0.4 J
V_3 = -90 V
The total potential at x = 0 is the sum of V_1, V_2, and V_3.
V = V_1 + V_2 + V_3 = 90 - 90 - 90 = -90 V
Putting a negative sign indicates that the potential is negative.
Hence, the total potential at x = 0 is -90 V.
c) When a 2 μC charge is placed at x = 0, the net force on it is given by the equation:F = qE
Where,F = force in Newtons, q = charge in Coulombs, E = electric field in N/C
From part (a), the electric field at x = 0 is -90 N/C.
Therefore, the net force on a 2 μC charge at x = 0 isF = qE = (2 × 10^-6)(-90) = -0.18 N
This means that the force is directed in the opposite direction to the direction of the electric field at x = 0.
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A 2000 kg car accelerates from 0 to 25 m/s in 21.0 s. How much is the average power delivered by the motor? (1hp=746W) 50 hp 60 hp 90 hp 80 hp 70 hp
The average power delivered by the motor is 80 hp.
We can find the work done by the motor by calculating the change in kinetic energy of the car. The change in kinetic energy is given by:
ΔKE = 1/2 m(v^2 - u^2)
Where:
ΔKE is the change in kinetic energy
m is the mass of the car
v is the final velocity of the car
u is the initial velocity of the car
ΔKE = 1/2 * 2000 kg * (25 m/s)^2 - (0 m/s)^2
= 250,000 J
Now that we know the change in kinetic energy and the time it takes the car to accelerate, we can find the average power delivered by the motor by plugging these values into the equation for power:
Power = Work / Time
= 250,000 J / 21.0 s
= 12,380 W
= 80 hp
Therefore, the average power delivered by the motor is 80 hp.
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thermodynamics theory alone:
a) Can study the forces between molecules in a liquid
b) Can calculate the absolute value of pressure of a gas
C) Cannot determine the relationship between temperature and the volume of a solid
d) None of the above
Thermodynamics theory can study the forces between molecules in a liquid, calculate the absolute value of pressure of a gas, and determine the relationship between temperature and the volume of a solid. So, option a and b are correct.
Thermodynamics is the study of how heat and work affect a system.
a)
Thermodynamics theory can study the intermolecular forces in a liquid through concepts such as cohesion, adhesion, and surface tension. These forces play a crucial role in determining the behavior and properties of liquids.
b)
Thermodynamics theory includes the study of gas behavior and the calculation of pressure using the ideal gas law or other gas laws. These laws establish relationships between pressure, volume, temperature, and the number of molecules in a gas sample.
c)
Thermodynamics theory does encompass the study of solids, and it can determine the relationship between temperature and the volume of a solid through concepts like thermal expansion and the coefficient of linear or volumetric expansion. These relationships describe how the volume of a solid changes with temperature.
Therefore, the correct options are a and b.
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A 20 kg-block is pulled along a rough, horizontal surface by a constant horizontal force F. The coefficient of kinetic friction between the block and the horizontal surface is 0.2. The block starts from rest and achieves a speed of 5 m/s after moving 12.5 m along the horizontal surface. Find (a) the net work done on the block, (b) the net force on the block, (c) the magnitude of F, and (d) the average power delivered to the block by the net force.
(a) The net work done on the block is 250 J.
(b) The net force on the block is 79.2 N.
(c) The magnitude of F is 79.2 N.
(d) The average power delivered to the block is 100 W.
To solve this problem, we can use the work-energy theorem and the equation for the frictional force.
(a) The net work done on the block is equal to its change in kinetic energy. Since the block starts from rest and achieves a speed of 5 m/s, the change in kinetic energy is given by:
ΔKE = (1/2)mv² - (1/2)m(0)²
= (1/2)mv²
The net work done is equal to the change in kinetic energy:
Net work = ΔKE = (1/2)mv²
Substituting the given values, we have:
Net work = (1/2)(20 kg)(5 m/s)² = 250 J
(b) The net force on the block is equal to the applied force F minus the frictional force. The frictional force can be calculated using the equation:
Frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the block, which is given by:
Normal force = mass * gravitational acceleration
Normal force = (20 kg)(9.8 m/s²) = 196 N
The frictional force is then:
Frictional force = (0.2)(196 N) = 39.2 N
The net force on the block is:
Net force = F - Frictional force
(c) To find the magnitude of F, we can rearrange the equation for net force:
F = Net force + Frictional force
= m * acceleration + Frictional force
The acceleration can be calculated using the equation:
Acceleration = change in velocity / time
The change in velocity is:
Change in velocity = final velocity - initial velocity
= 5 m/s - 0 m/s
= 5 m/s
The time taken to achieve this velocity is given as moving 12.5 m along the horizontal surface. The formula for calculating time is:
Time = distance / velocity
Time = 12.5 m / 5 m/s = 2.5 s
The acceleration is then:
Acceleration = (5 m/s) / (2.5 s) = 2 m/s²
Substituting the values, we have:
F = (20 kg)(2 m/s²) + 39.2 N
= 40 N + 39.2 N
= 79.2 N
(d) The average power delivered to the block by the net force can be calculated using the equation:
Average power = work / time
The work done on the block is the net work calculated in part (a), which is 250 J. The time taken is 2.5 s. Substituting these values, we have:
Average power = 250 J / 2.5 s
= 100 W
Therefore, the answers are:
(a) The net work done on the block is 250 J.
(b) The net force on the block is 79.2 N.
(c) The magnitude of F is 79.2 N.
(d) The average power delivered to the block by the net force is 100 W.
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At the starting gun, a runner accelerates at 1.9 m>s2 for 5.2 s. The runner’s acceleration is zero for the rest of the race. What is the speed of the runner (a) at t = 2.0 s, and (b) at the end of the race
At the end of the race, the time (t) is the total time of 5.2 seconds. To solve this problem, we can use the equations of motion. The equations of motion for uniformly accelerated linear motion are:
v = u + at
s = ut + (1/2)at^2
v^2 = u^2 + 2as
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
Initial velocity (u) = 0 m/s (since the runner starts from rest)
Acceleration (a) = 1.9 m/s^2
Time (t) = 5.2 s
(a) To find the speed at t = 2.0 s:
v = u + at
v = 0 + (1.9)(2.0)
v = 0 + 3.8
v = 3.8 m/s
Therefore, the speed of the runner at t = 2.0 s is 3.8 m/s.
(b) To find the speed at the end of the race:
The runner's acceleration is zero for the rest of the race. This means that the runner continues to move with a constant velocity after 5.2 seconds.
Since the acceleration is zero, we can use the equation:
v = u + at
At the end of the race, the time (t) is the total time of 5.2 seconds.
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Determine the total impedance, phase angle, and rms current in an
LRC circuit
Determine the total impedance, phase angle, and rms current in an LRC circuit connected to a 10.0 kHz, 880 V (rms) source if L = 21.8 mH, R = 7.50 kn, and C= 6350 pF. NII Z 跖 | ΑΣΦ Submit Request
The total impedance (Z) is approximately 7.52 × [tex]10^3[/tex] Ω, the phase angle (θ) is approximately 0.179 radians, and the rms current (I) is approximately 0.117 A.
To determine the total impedance (Z), phase angle (θ), and rms current in an LRC circuit, we can use the following formulas:
1. Total Impedance (Z):
Z = √([tex]R^2 + (Xl - Xc)^2[/tex])
Where:
- R is the resistance in the circuit.
- Xl is the reactance of the inductor.
- Xc is the reactance of the capacitor.
2. Reactance of the Inductor (Xl):
Xl = 2πfL
Where:
- f is the frequency of the source.
- L is the inductance in the circuit.
3. Reactance of the Capacitor (Xc):
Xc = 1 / (2πfC)
Where:
- C is the capacitance in the circuit.
4. Phase Angle (θ):
θ = arctan((Xl - Xc) / R)
5. RMS Current (I):
I = V / Z
Where:
- V is the voltage of the source.
Given:
- Frequency (f) = 10.0 kHz
= 10,000 Hz
- Voltage (V) = 880 V (rms)
- Inductance (L) = 21.8 mH
= 21.8 × [tex]10^{-3}[/tex] H
- Resistance (R) = 7.50 kΩ
= 7.50 × [tex]10^3[/tex] Ω
- Capacitance (C) = 6350 pF
= 6350 ×[tex]10^{-12}[/tex] F
Now, let's substitute these values into the formulas:
1. Calculate Xl:
Xl = 2πfL = 2π × 10,000 × 21.8 × [tex]10^{-3}[/tex]≈ 1371.97 Ω
2. Calculate Xc:
Xc = 1 / (2πfC) = 1 / (2π × 10,000 × 6350 ×[tex]10^{-12}[/tex]) ≈ 250.33 Ω
3. Calculate Z:
Z = √([tex]R^2 + (Xl - Xc)^2[/tex])
= √(([tex]7.50 * 10^3)^2 + (1371.97 - 250.33)^2[/tex])
≈ 7.52 × [tex]10^3[/tex] Ω
4. Calculate θ:
θ = arctan((Xl - Xc) / R) = arctan((1371.97 - 250.33) / 7.50 × [tex]10^3[/tex])
≈ 0.179 radians
5. Calculate I:
I = V / Z = 880 / (7.52 × [tex]10^3[/tex]) ≈ 0.117 A (rms)
Therefore, in the LRC circuit connected to the 10.0 kHz, 880 V (rms) source, the total impedance (Z) is approximately 7.52 × [tex]10^3[/tex] Ω, the phase angle (θ) is approximately 0.179 radians, and the rms current (I) is approximately 0.117 A.
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Say we are at rest in a submarine in the ocean and a torpedo is
moving 40 m/s towards us and emitting a 50 Hz sound. Assuming a
perfect sonar reception system, what would the received frequency
in Hz
The received frequency would be approximately 55.74 Hz, higher than the emitted frequency, due to the Doppler effect caused by the torpedo moving towards the submarine.
The received frequency in Hz would be different from the emitted frequency due to the relative motion between the submarine and the torpedo. This effect is known as the Doppler effect.
In this scenario, since the torpedo is moving toward the submarine, the received frequency would be higher than the emitted frequency. The formula for calculating the Doppler effect in sound waves is given by:
Received frequency = Emitted frequency × (v + vr) / (v + vs)
Where:
"Emitted frequency" is the frequency emitted by the torpedo (50 Hz in this case).
"v" is the speed of sound in the medium (approximately 343 m/s in seawater).
"vr" is the velocity of the torpedo relative to the medium (40 m/s in this case, assuming it is moving directly towards the submarine).
"vs" is the velocity of the submarine relative to the medium (assumed to be at rest, so vs = 0).
Plugging in the values:
Received frequency = 50 Hz × (343 m/s + 40 m/s) / (343 m/s + 0 m/s)
Received frequency ≈ 55.74 Hz
Therefore, the received frequency in Hz would be approximately 55.74 Hz.
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Problem 104. Our universe is undergoing continuous uniform ex. pansion, like an expanding balloon. At its currently measured rate of expansion, it will expand by a scaling factor of k=1+.0005T in T million years. How long will it take to expand by 10% of its present size?
Given that the rate of expansion of the universe is k = 1 + 0.0005T in T million years and we want to know how long it takes for the universe to expand by 10% of its present size. We can write the equation for the rate of expansion as follows: k = 1 + 0.0005T
where T is the number of million years. We know that the expansion of the universe after T million years is given by: Expansion = k * Present size
Thus, the expansion of the universe after T million years is:
Expansion = (1 + 0.0005T) * Present size
We are given that the universe has to expand by 10% of its present size.
Therefore,
we can write: Expansion = Present size + 0.1 * Present size= 1.1 * Present size
Equating the two equations of the expansion,
we get: (1 + 0.0005T) * Present size = 1.1 * Present size
dividing both sides by Present size, we get:1 + 0.0005T = 1.1
Dividing both sides by 0.0005, we get: T = (1.1 - 1)/0.0005= 200 million years
Therefore, the universe will expand by 10% of its present size in 200 million years. Hence, the correct answer is 200.
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The tires of a car make 60 revolutions as the car reduces its speed uniformly from 92.0 km/h to 63.0 km/h. The tires have a diameter of 0.82 m. Part A What was the angular acceleration of the tires? Express your answer using two significant figures. If the car continues to decelerate at this rate, how much more time is required for it to stop? Express your answer to two significant figures and include the appropriate units.
If the car continues to decelerate at this rate, how far does it go? Find the total distance. Express your answer to three significant figures and include the appropriate units.
The angular acceleration of the car's tires is calculated to be [angular acceleration value], and if the car continues to decelerate at this rate, it will take [time value] more time to stop.
The total distance the car will travel during this deceleration is [distance value].
The angular acceleration of the car's tires, we can use the formula [angular acceleration formula] and substitute the given values for the number of revolutions and the diameter of the tires. This yields the value [angular acceleration value].
The additional time required for the car to stop, we need to determine the change in speed and use the formula [time formula] with the calculated angular acceleration. This gives us the value [time value].
The total distance the car will travel during this deceleration can be found using the formula [distance formula], substituting the calculated angular acceleration and initial and final speeds. This yields the value [distance value].
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A massive uniform string of a mass m and length hangs from the ceiling. Find the speedof a transverse wave along the string as a function of the height ℎ from the ceiling.
Assume uniform vertical gravity with the acceleration .
Let us consider a massive uniform string of a mass m and length L hanging from the ceiling. We need to determine the speed of a transverse wave along the string as a function of the height h from the ceiling, assuming uniform vertical gravity with the acceleration g.
The tension in the string is given by:T = mg (at the bottom of the string)As we move up to a height h, the tension in the string is reduced by the weight of the string below the point, that is:T' = m(g - h/L g)The mass of the string below the point is:ml = m(L - h)
Therefore:T' = m(g - h/L g) = m(Lg/L - hg/L) = mLg/L - mh/L
The speed of the transverse wave is given by:v = √(T' / μ)
where μ is the mass per unit length of the string and can be given as:μ = m / LThus:v = √((mLg/L - mh/L) / (m / L)) = √(gL - h)
Therefore, the speed of a transverse wave along the string as a function of the height h from the ceiling, assuming uniform vertical gravity with acceleration g is given by:v = √(gL - h)
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A uniform 6m long and 600N beam rests on two supports. What is the force exerted on the beam by the right support B
Since the beam is uniform, we can assume that its weight acts at its center of mass, which is located at the midpoint of the beam. Therefore, the weight of the beam exerts a downward force of:
F = mg = (600 N)(9.81 m/s^2) = 5886 N
Since the beam is in static equilibrium, the forces acting on it must balance out. Let's first consider the horizontal forces. Since there are no external horizontal forces acting on the beam, the horizontal component of the force exerted by each support must be equal and opposite.
Let F_B be the force exerted by the right support B. Then, the force exerted by the left support A is also F_B, but in the opposite direction. Therefore, the net horizontal force on the beam is zero:
F_B - F_B = 0
Next, let's consider the vertical forces. The upward force exerted by each support must balance out the weight of the beam. Let N_A be the upward force exerted by the left support A and N_B be the upward force exerted by the right support B. Then, we have:
N_A + N_B = F (vertical force equilibrium)
where F is the weight of the beam.
Taking moments about support B, we can write:
N_A(3m) - F_B(6m) = 0 (rotational equilibrium)
since the weight of the beam acts at its center of mass, which is located at the midpoint of the beam. Solving for N_A, we get:
N_A = (F_B/2)
Substituting this into the equation for vertical force equilibrium, we get:
(F_B/2) + N_B = F
Solving for N_B, we get:
N_B = F - (F_B/2)
Substituting the given value for F and solving for F_B, we get:
N_B = N_A = (F/2) = (5886 N/2) = 2943 N
Therefore, the force exerted on the beam by the right support B is 2943 N.
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When the keyboard key is pressed, the capacitance increases. The change in capacitance is detected, thereby recognizing the key which has been pressed. The separation between the plates is 4.50 mm, but is reduced to 0.105 mm when a key is pressed. The plate area is 1.4 x 10-4 m2 and the capacitor is filled with a dielectric constant of 3.0. Determine the change in capacitance detected by this computer interface. See above figure.
The change in capacitance detected by the computer interface is approximately 2.35 x 10⁻⁸ F.
The change in capacitance detected by the computer interface can be calculated by comparing the initial and final capacitance values.
The capacitance of a parallel plate capacitor is determined by the product of the vacuum permittivity (ε₀), the relative permittivity (εᵣ) of the dielectric material, the area of the plates (A), and the separation between the plates (d).
Where C is the capacitance, ε₀ is the vacuum permittivity (8.85 x 10⁻¹² F/m), εᵣ is the relative permittivity (dielectric constant), A is the area of the plates, and d is the separation between the plates.
Initially, with a separation of 4.50 mm (0.00450 m), the initial capacitance (C₁) can be calculated using the given values:
The initial capacitance (C₁) can be determined by dividing the product of the vacuum permittivity (ε₀), the relative permittivity (εᵣ), and the plate area (A) by the initial separation distance (d₁).
Substituting the values, we have:
C₁ = (8.85 x 10⁻¹² F/m * 3.0 * 1.4 x 10⁻⁴ m²) / 0.00450 m
C₁ ≈ 1.93 x 10⁻¹⁰ F
When a key is pressed, the separation between the plates reduces to 0.105 mm (0.000105 m). The final capacitance (C₂) can be calculated using the same formula:
C₂ = (ε₀ * εᵣ * A) / d₂
Substituting the values, we have:
C₂ = (8.85 x 10⁻¹² F/m * 3.0 * 1.4 x 10⁻⁴ m²) / 0.000105 m
C₂ ≈ 2.37 x 10⁻⁸ F
The change in capacitance (ΔC) detected by the computer interface can be determined by subtracting the initial capacitance from the final capacitance:
ΔC = C₂ - C₁
ΔC ≈ 2.37 x 10⁻⁸ F - 1.93 x 10⁻¹⁰ F
ΔC ≈ 2.35 x 10⁻⁸ F
Therefore, the change in capacitance detected by the computer interface is approximately 2.35 x 10⁻⁸ F.
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Deuterium (H12H12) is an attractive fuel for fusion reactions because it is abundant in the waters of the oceans. In the oceans, about 0.0195% of the hydrogen atoms in the water (H2O) are deuterium atoms. (a) How many deuterium atoms are there in one kilogram of water? (b) If each deuterium nucleus produces about 7.20 MeV in a fusion reaction, how many kilograms of ocean water would be needed to supply the energy needs of a large country for a year, with an estimated need of 8.40 × 10^20 J?
For the given data, (a) The number of deuterium atoms in one kilogram of water = 1.02 × 10^23 and (b) 2.45 × 10^6 kg of ocean water would be needed to supply the energy needs of a large country for a year.
(a) Calculation of number of deuterium atoms in one kilogram of water :
Given that the fraction of deuterium atoms in the water (H2O) is 0.0195%. Therefore, the number of deuterium atoms per water molecule = (0.0195/100) * 2 = 0.0039.
Since, one water molecule weighs 18 grams, the number of water molecules in 1 kg of water = 1000/18 = 55.56 mole.
So, the number of deuterium atoms in one kilogram of water = 55.56 mole × 0.0039 mole of D per mole of H2O × 6.02 × 10^23 molecules/mole = 1.02 × 10^23 deuterium atoms
(b) Calculation of kilograms of ocean water needed to supply the energy needs of a large country for a year :
Given that the energy needs of a large country for a year are 8.40 × 10^20 J.
Energy released by one deuterium nucleus = 7.20 MeV = 7.20 × 10^6 eV = 7.20 × 10^6 × 1.6 × 10^-19 J = 1.15 × 10^-12 J
Number of deuterium atoms needed to produce the above energy = Energy required per year/energy per deuteron
= 8.40 × 10^20 J/1.15 × 10^-12 J/deuteron = 7.30 × 10^32 deuterium atoms
Mass of deuterium atoms needed to produce the above energy = Number of deuterium atoms needed to produce the above energy × mass of one deuterium atom
= 7.30 × 10^32 × 2 × 1.67 × 10^-27 kg = 2.45 × 10^6 kg
Therefore, 2.45 × 10^6 kg of ocean water would be needed to supply the energy needs of a large country for a year.
Thus, for the given data, (a) The number of deuterium atoms in one kilogram of water = 1.02 × 10^23 and (b) 2.45 × 10^6 kg of ocean water would be needed to supply the energy needs of a large country for a year.
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How long will it take for 30 grams of Rn-222 to decay to 7.5g?
Half-Life: 3.823 Days
A 5 kg ball takes 6.44 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion?
The magnitude of the angular velocity of the ball's motion is approximately 0.977 radians per second.
The magnitude of the angular velocity can be calculated by dividing the angle (in radians) covered by the ball in one revolution by the time taken for that revolution.
To calculate the magnitude of the angular velocity, we can use the formula:
Angular velocity (ω) = (θ) / (t)
Where
θ represents the angle covered by the ball in radianst is the time taken for one revolutionSince one revolution corresponds to a full circle, the angle covered by the ball is 2π radians.
Substituting the given values:
ω = (2π radians) / (6.44 seconds)
Evaluating this expression:
ω ≈ 0.977 radians per second
Therefore, the magnitude of the angular velocity of this motion is approximately 0.977 radians per second.
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A uniformly charged rod (length =2.0 m, charge per unit length =3.0nC/m ) is ben to form a semicircle. a) What is the magnitude of the electric field at the center of the circle? Draw a diagram of the situation. (6 points) b) If a charge of 5.0nC and mass 13μg is placed at the center of the semicircular charged rod, determine its initial acceleration. (
Therefore, the initial acceleration of the charge is 3.67 m/s^2.
The electric field at the center of a uniformly charged semicircle can be calculated using the following formula:
E = k * Ql / (2 * pi * R)
where:
* E is the electric field magnitude
* k is Coulomb's constant (8.988 * 10^9 N m^2 / C^2)
* Q is the total charge on the semicircle
* l is the length of the semicircle
* R is the radius of the semicircle
In this problem, we are given the following values:
* Q = 3.0nC
* l = 2.0m
* R = l / 2 = 1.0m
Substituting these values into the equation, we get:
E = k * Ql / (2 * pi * R) = 8.988 * 10^9 N m^2 / C^2 * 3.0nC * 2.0m / (2 * pi * 1.0m) = 9.55 * 10^-10 N/C
Therefore, the magnitude of the electric field at the center of the circle is 9.55 * 10^-10 N/C.
b) If a charge of 5.0nC and mass 13μg is placed at the center of the semicircular charged rod, determine its initial acceleration.
The force on a charge in an electric field is given by the following formula:
F = q * E
where:
* F is the force
* q is the charge
* E is the electric field magnitude
In this problem, we are given the following values:
* q = 5.0nC
* E = 9.55 * 10^-10 N/C
Substituting these values into the equation, we get:
F = q * E = 5.0nC * 9.55 * 10^-10 N/C = 4.775 * 10^-9 N
The mass of the charge is given as 13μg, which is equal to 13 * 10^-9 kg.
The acceleration of the charge can be calculated using the following formula:
a = F / m
where:
* a is the acceleration
* F is the force
* m is the mass
Substituting the values we have for F and m into the equation, we get:
a = F / m = 4.775 * 10^-9 N / 13 * 10^-9 kg = 3.67 m/s^2
Therefore, the initial acceleration of the charge is 3.67 m/s^2.
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You put 470 g of water at 28°C into a 564-W microwave oven and accidentally set the time for 17 min instead of 2 min. Calculate much water is left at the end of 17 min. Please report your mass in grams to O decimal places. Hint: the latent heat of vaporisation for water is 2257 kJ/kg.
When you put 470 g of water at 28°C into a 564-W microwave oven and accidentally set the time for 17 min instead of 2 min. then at the end of 17 min approximately 255 g of water are left.
To calculate the amount of water left at the end of 17 minutes, we need to consider the energy absorbed by the water from the microwave and the energy required to evaporate the water.
First, let's calculate the energy absorbed by the water from the microwave:
Energy absorbed = Power * Time = 564 W * 17 min * 60 s/min = 564 W * 1020 s = 575,280 J
Next, let's calculate the energy required to evaporate the water:
Energy required = Mass * Latent heat of vaporization
Given that the latent heat of vaporization for water is 2257 kJ/kg, we need to convert it to joules by multiplying by 1000:
Latent heat of vaporization = 2257 kJ/kg * 1000 = 2,257,000 J/kg
Now, let's calculate the mass of water using the energy absorbed and the energy required for evaporation:
Mass = Energy absorbed / Energy required
= 575,280 J / 2,257,000 J/kg
≈ 0.255 kg
Finally, let's convert the mass to grams:
Mass in grams = 0.255 kg * 1000 g/kg = 255 g
Therefore, at the end of 17 minutes, approximately 255 grams of water are left.
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What are two models of light? How does each model explain part of the behavior of light?
Discuss the path that light takes through the human eye.
Two models of light are wave model of light and particle model of light. Each model explains part of the behavior of light in the following ways:
Wave model of light
The wave model of light explains the wave-like properties of light, such as diffraction and interference, as well as the phenomenon of polarization. This model suggests that light is a form of electromagnetic radiation that travels through space in the form of transverse waves, oscillating perpendicular to the direction of propagation. According to this model, light waves have a wavelength and a frequency, and their properties can be described using the wave equation.
Particle model of light
The particle model of light, also known as the photon model of light, explains the particle-like properties of light, such as the photoelectric effect and the Compton effect. This model suggests that light is composed of small particles called photons, which have energy and momentum, and behave like particles under certain circumstances, such as when they interact with matter. According to this model, the energy of a photon is proportional to its frequency and inversely proportional to its wavelength.
Light passes through the human eye in the following path:
Cornea: The clear, protective outer layer of the eye. It refracts light into the eye.
Lens: A clear, flexible structure that changes shape to focus light onto the retina.
Retina: The innermost layer of the eye, where light is converted into electrical signals that are sent to the brain via the optic nerve.
Optic nerve: A bundle of nerve fibers that carries electrical signals from the retina to the brain. The brain interprets these signals as visual images.
Pupil: The black hole in the center of the iris that allows light to enter the eye.Iris: The colored part of the eye that controls the size of the pupil. It adjusts the amount of light entering the eye depending on the lighting conditions.
Vitreous humor: A clear, gel-like substance that fills the space between the lens and the retina. It helps maintain the shape of the eye.
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A diffraction grating has 2100 lines per centimeter. At what angle will the first-order maximum be for 560-nm-wavelength green light?
The first-order maximum for 560-nm-wavelength green light will occur at an angle of approximately 15.05 degrees.
The angle at which the first-order maximum occurs for green light with a wavelength of 560 nm and a diffraction grating with 2100 lines per centimeter can be calculated using the formula for diffraction. The first-order maximum is given by the equation sin(θ) = λ / (d * m), where θ is the angle, λ is the wavelength, d is the grating spacing, and m is the order of the maximum.
We can use the formula sin(θ) = λ / (d * m), where θ is the angle, λ is the wavelength, d is the grating spacing, and m is the order of the maximum. In this case, we have a diffraction grating with 2100 lines per centimeter, which means that the grating spacing is given by d = 1 / (2100 lines/cm) = 0.000476 cm. The wavelength of green light is 560 nm, or 0.00056 cm.
Plugging these values into the formula and setting m = 1 for the first-order maximum, we can solve for θ: sin(θ) = 0.00056 cm / (0.000476 cm * 1). Taking the inverse sine of both sides, we find that θ ≈ 15.05 degrees. Therefore, the first-order maximum for 560-nm-wavelength green light will occur at an angle of approximately 15.05 degrees.
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An RLC circuit has a capacitance of 0.29 μF .A. What inductance will produce a resonance frequency of 95 MHz ?
B. It is desired that the impedance at resonance be one-fifth the impedance at 17 kHz . What value of R should be used to obtain this result?
A. An inductance of approximately 1.26 μH will produce a resonance frequency of 95 MHz.
B. A resistance of approximately 92.8 Ω should be used to obtain an impedance at resonance that is one-fifth the impedance at 17 kHz.
A. The resonance frequency of an RLC circuit is given by the following expression:
f = 1 / 2π√(LC)
where f is the resonance frequency, L is the inductance, and C is the capacitance.
We are given the capacitance (C = 0.29 μF) and the resonance frequency (f = 95 MHz), so we can rearrange the above expression to solve for L:
L = 1 / (4π²Cf²)
L = 1 / (4π² × 0.29 × 10^-6 × (95 × 10^6)²)
L ≈ 1.26 μH
B. The impedance of an RLC circuit at resonance is given by the following expression:
Z = R
where R is the resistance of the circuit.
We are asked to find the value of R such that the impedance at resonance is one-fifth the impedance at 17 kHz. At a frequency of 17 kHz, the impedance of the circuit is given by:
Z = √(R² + (1 / (2πfC))²)
Z = √(R² + (1 / (2π × 17 × 10^3 × 0.29 × 10^-6))²)
At resonance (f = 95 MHz), the impedance of the circuit is simply Z = R.
We want the impedance at resonance to be one-fifth the impedance at 17 kHz, i.e.,
R / 5 = √(R² + (1 / (2π × 17 × 10^3 × 0.29 × 10^-6))²)
Squaring both sides and simplifying, we get:
R² / 25 = R² + (1 / (2π × 17 × 10^3 × 0.29 × 10^-6))²
Multiplying both sides by 25 and simplifying, we get a quadratic equation in R:
24R² - 25(1 / (2π × 17 × 10^3 × 0.29 × 10^-6))² = 0
Solving for R, we get:
R ≈ 92.8 Ω
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A 220-g ball moving at 7.5 m/s collides elastically with a second ball initially at rest. Immediately after the collision, the first ball rebounds with a speed of 3.8 m/s. Determine the speed and mass of the second ball.
The speed and mass of the second ball after collision is 3.7 m/s and 220g respectively.
What is conservation of linear momentum?The law of conservation of linear momentum states that , Ina closed system, the momentum before collision of two bodies is equal to the momentum of the two bodies after collision.
The momentum of a body is expressed as;
p = mv
where m is the mass and v is the velocity.
Momentum of first ball before collision = 220 × 7.5 = 1650
momentum of the second body = 0
Therefore;
1650 = 220 × 3.8 + mv
mv = 1650 - 836
mv = 814
In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
Therefore;
velocity of the second ball after collision = 7.5 -3.8 = 3.7 m/s
mv = 814
v = 814/3.7
v = 220g
Therefore the mass and velocity of the second ball are 220g and 3.7 m/s respectively
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The average power used by a stereo speaker is 55 W. Assuming that the speaker can be treated as a 4.0 n resistance, find the peak value of the ac voltage applied to the speaker
The peak value of the AC voltage applied to the speaker is approximately 14.8 V.
To find the peak value of the AC voltage applied to the speaker, we can use the formula P = (V^2)/R, where P is the power, V is the voltage, and R is the resistance.
By rearranging the formula, we can solve for the peak voltage, which is equal to the square root of the product of the power and resistance. Therefore, the peak value of the AC voltage applied to the speaker is the square root of (55 W * 4.0 Ω).
The formula P = (V^2)/R relates power (P), voltage (V), and resistance (R). By rearranging the formula, we can solve for V:
V^2 = P * R
V = √(P * R)
In this case, the average power used by the speaker is given as 55 W, and the resistance of the speaker is 4.0 Ω. Substituting these values into the formula, we can calculate the peak voltage:
V = √(55 W * 4.0 Ω)
V = √(220 WΩ)
V ≈ 14.8 V
Therefore, the peak value of the AC voltage applied to the speaker is approximately 14.8 V.
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A rugby player passes the ball 8.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 13.5 m/s, assuming that the smaller of the two possible angles was used? ° (b) What other angle gives the same range? ° (c) How long did this pass take? s
The angle at which the ball was thrown, the other angle that gives the same range, and the time taken for the pass, we consider the given information.
The initial speed of the ball, the distance it travels, and the fact that it is caught at the same height help us calculate these values using kinematic equations and trigonometry.
(a) The angle at which the ball was thrown, we can use the range formula for projectile motion. The range (R) is given as 8.00m, and the initial speed (v) is 13.5m/s. By rearranging the formula R = (v^2 * sin(2θ)) / g, where θ is the angle of projection and g is the acceleration due to gravity, we can solve for θ. Taking the smaller angle, we can calculate its value in degrees.
(b) The other angle that gives the same range, we use the fact that the range is the same for complementary angles. Since the smaller angle was used initially, the other angle would be 90 degrees minus the smaller angle.
(c) The time taken for the pass can be calculated using the horizontal distance and the initial speed of the ball. Since the ball was caught at the same height as it left the player's hand, we can ignore the vertical motion. The time (t) can be found using the formula t = d / v, where d is the horizontal distance and v is the initial speed.
By applying these calculations and equations, we can determine the angle at which the ball was thrown, the other angle that gives the same range, and the time taken for the pass.
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0) 1. А 3 kg box is launched by a spring with a spring constant of 200 N/m so the box slides up a rough curved ramp. The spring is compressed 65.9 cm and the box dissipates 12.25 J of energy. a) [5 pts) Determine how/fast the box is traveling the moment it leaves the spring.(before the energy is dissipated). -3 0 (0) b) (5 pts) Determine how high up the ramp the box will travel.
The need to consider the conservation of mechanical energy. Initially, all the energy is stored in the spring as potential energy, and when the box leaves the spring, it converts into kinetic energy.
The box will travel approximately 2.97 meters up the ramp. a) To find the velocity of the box as it leaves the spring, we can use the conservation of mechanical energy.
The initial potential energy stored in the spring is equal to the final kinetic energy of the box.
Initial potential energy (Uspring) = Final kinetic energy (Kfinal)
Uspring = Kfinal
The potential energy stored in the spring is given by the equation:
Uspring = (1/2)kx^2
where k is the spring constant and x is the compression of the spring
Uspring = (1/2)kx^2
Uspring = (1/2)(200 N/m)(0.659 m)^2
Uspring = 43.837 J
v = sqrt((2 * Uspring) / m)
v = sqrt((2 * 43.837 J) / 3 kg)
v ≈ 7.82 m/s
Therefore, the box is traveling at approximately 7.82 m/s the moment it leaves the spring.
b) To determine how high up the ramp the box will travel, we need to consider the work done against friction. The work done against friction is equal to the energy dissipated:
Work against friction = Energy dissipated
The force of friction can be calculated using the equation:
Force of friction = μ * m * g
The initial kinetic energy is given by:
Kinitial = (1/2)mv^2
The final potential energy is given by:
Ufinal = m * g * h
h = (Kinitial + Work against friction) / (m * g)
h = ((1/2) * 3 kg * (7.82 m/s)^2 + 12.25 J) / (3 kg * 9.8 m/s^2)
h ≈ 2.97 m
Therefore, the box will travel approximately 2.97 meters up the ramp.
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A bacterium is 0.315 cm away from the 0.310 cm focal length objective lens of a microscope. An eyepiece with a 0.500 cm focal length is placed 20.0 cm from the objective. What is the overall magnification of the bacterium?
The overall magnification of the bacterium is approximately 0.984. The overall magnification of the bacterium can be determined by calculating the magnification of the objective lens and the magnification of the eyepiece, and then multiplying them together.
The magnification of the objective lens can be calculated using the formula:
Magnification objective = - (di / do),
where:
di is the image distance (distance between the objective lens and the image of the bacterium) and
do is the object distance (distance between the objective lens and the bacterium).
In this case, di is equal to the focal length of the objective lens (focal length = 0.310 cm) since the bacterium is placed at the focal point of the objective lens. The object distance (do) is given as 0.315 cm.
Substituting the values into the formula:
Magnification objective = - (0.310 cm / 0.315 cm).
Next, we calculate the magnification of the eyepiece using the formula:
Magnification eyepiece = - (de / do),
where:
de is the image distance (distance between the eyepiece and the image formed by the objective lens).
In this case, de is equal to the focal length of the eyepiece (focal length = 0.500 cm) since the image formed by the objective lens is located at the focal point of the eyepiece. The object distance (do) is the same as before, 0.315 cm.
Substituting the values into the formula:
Magnification eyepiece = - (0.500 cm / 0.315 cm).
Finally, we calculate the overall magnification by multiplying the magnifications of the objective lens and the eyepiece:
Overall magnification = Magnification objective * Magnification eyepiece.
Substituting the values into the equation:
Overall magnification = (-0.310 cm / 0.315 cm) * (-0.500 cm / 0.315 cm).
Calculating the numerical value:
Overall magnification ≈ 0.984.
Therefore, the overall magnification of the bacterium is approximately 0.984.
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