When light travels from air into glass, it refracts towards the normal at the air-glass interface, propagates through the glass, and upon exiting, refracts away from the normal back into the air.
When light travels from air into glass, it undergoes several changes in its path due to the difference in optical properties between the two mediums. The path light takes can be described as follows:
1. Incident Ray: The journey begins with an incident ray, which is the incoming light ray traveling through the air towards the glass surface.
2. Refraction: As the incident ray reaches the interface between air and glass, it encounters a change in the refractive index. Refractive index is a measure of how much a medium can bend light. Glass has a higher refractive index than air, so the incident ray bends towards the normal, which is an imaginary line perpendicular to the surface of the glass.
3. Transmission: After refraction, the incident ray enters the glass and continues its path through the medium. Inside the glass, the light travels in a straight line until it encounters another interface or is affected by other optical phenomena.
4. Internal Reflection: If the incident ray encounters a glass-air interface at an angle greater than the critical angle, total internal reflection can occur. In this case, the light reflects back into the glass instead of transmitting out, effectively bouncing off the interface.
5. Refraction (again): If the incident ray does not undergo total internal reflection, it continues to propagate through the glass. At another glass-air interface, the light exits the glass and enters the air again. This transition involves refraction once more, but this time the light bends away from the normal, returning to its original path in air.
6. Transmitted Ray: Finally, the light ray continues to travel through the air, maintaining its original direction and path as it moves away from the glass surface.
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A spring oriented vertically is attached to a hard horizontal surface as in the figure below. The spring has a force constant of 1.30 kN/m. How much is the spring compressed when a object of mass m = 2.70 kg is placed on top of the spring and the system is at rest? Answer should be in centimeters.
The spring is compressed by approximately 2.04 cm. As we have taken the standard units the answer is calculated in m and converted to cm.
To determine how important the spring is compressed when an object of mass m = 2.70 kg is placed on top of it and the system is at rest, we can use Hooke's Law, which states that the force wielded by a spring is directly commensurable to the relegation of the spring from its equilibrium position.
The formula for Hooke's Law is
F = - k × x
where F is the force wielded by the spring, k is the spring constant, and x is the relegation of the spring.
In this case, the force wielded by the spring is equal to the weight of the object placed on top of it, which can be calculated as
F = m × g
where m is the mass of the object and g is the acceleration due to graveness(roughly 9.8 m/ s²).
Given
Mass( m ) = 2.70 kg
Spring constant( k) = 1.30 kN/ m( Note 1 kN = 1000 N)
Converting the spring constant to Newtons
k = 1.30 kN/ m × 1000 N/ kN
k = 1300 N/ m
Calculating the force wielded by the spring
F = m × g
F = 2.70 kg × 9.8 m/ s²
F ≈26.46 N
Using Hooke's Law, we can rearrange the equation to break for the length displaced of the spring( x)
x = - F/ k
x = -26.46 N/ 1300 N/ m
x ≈-0.0204 m
The negative sign indicates that the spring is compressed. thus, when the object of mass m = 2.70 kg is placed on top of the spring and the system is at rest.
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Plsss help Bumper car A (282 kg) moving +2.82 m/s
makes an elastic collision with bumper
car B (210 kg) moving +1.72 m/s. What is
the velocity of car A after the collision?
(Unit = m/s)
Remember: right is +, left is -
Answer:
Approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Let [tex]m_{A}[/tex] and [tex]m_{B}[/tex] denote the mass of the two vehicles. Let [tex]u_{A}[/tex] and [tex]u_{B}[/tex] denote the velocity before the collision. Let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] denote the velocity after the collision.
Since the collision is elastic, both momentum and kinetic energy should be conserved.
For momentum to conserve:
[tex]m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B}[/tex].
For kinetic energy to conserve:
[tex]\displaystyle \frac{1}{2}\, m_{A} \, ({v_{A}}^{2}) + \frac{1}{2}\, m_{B} \, ({v_{B}}^{2}) = \frac{1}{2}\, m_{A}\, ({u_{A}}^{2}) + \frac{1}{2}\, m_{B}\, ({u_{B}}^{2})[/tex].
Simplify to obtain:
[tex]\displaystyle m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})[/tex].
It is given that [tex]m_{A} = 282\; {\rm kg}[/tex], [tex]m_{B} = 210\; {\rm kg}[/tex], [tex]u_{A} = 2.82\; {\rm m\cdot s^{-1}}[/tex], and [tex]u_{B} = 1.72\; {\rm m\cdot s^{-1}}[/tex]. The value (in [tex]{\rm m\cdot s^{-1}}[/tex]) of [tex]v_{A}[/tex] and [tex]v_{B}[/tex] can be found by solving this nonlinear system of two equations and two unknowns:
[tex]\left\lbrace \begin{aligned} & m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B} \\ & m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})\end{aligned}\right.[/tex].
[tex]\left\lbrace \begin{aligned} & 282 \, v_{A} + 210 \, v_{B} = 282\, (2.82) + 210\, (1.72) \\ & 282 \, ({v_{A}}^{2}) + 210 \, ({v_{B}}^{2}) = 282\, ({2.82}^{2}) + 210\, ({1.72}^{2})\end{aligned}\right.[/tex].
Solving this system gives two possible sets of solutions:
[tex]\left\lbrace\begin{aligned}v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}} \\ v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}\end{aligned}\right.[/tex].[tex]\left\lbrace\begin{aligned}v_{A} &\approx 2.82\; {\rm m\cdot s^{-1}} \\ v_{B} &\approx 1.72\; {\rm m\cdot s^{-1}}\end{aligned}\right.[/tex].However, the second set of solutions is invalid since it suggests that the velocity of the two vehicles stayed unchanged after the collision. Hence, only the first set of solutions ([tex]v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}}[/tex], [tex]v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}[/tex]) is valid.
Therefore, the velocity of vehicle [tex]A[/tex] would be approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex] after the collision.
About 1.75% of water on Earth is in Greenland and Antarctica's icecaps, and about 97.5% is in the oceans. Assume the icecaps have an average temperature of -28°C, and the oceans have an average temperature of 4.8°C. If all the icecaps slid into the ocean and melted, how much would the average temperature of the ocean decrease?
If all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
To calculate the decrease in the average temperature of the ocean when all the icecaps melt, we need to consider the heat exchange between the icecaps and the ocean.
Let's start by calculating the heat released by the icecaps when they melt. We can use the specific heat capacity formula:
Heat released = Mass of icecaps × Specific heat capacity of ice × Temperature change
Since the icecaps constitute 1.75% of the Earth's water, the mass of icecaps is 0.0175 times the total mass of water on Earth.
Assuming the icecaps have an average temperature of -28°C and melt into liquid water at 0°C, the temperature change is 0°C - (-28°C) = 28°C.
Next, we need to calculate the heat absorbed by the ocean when the icecaps melt. Using the same formula:
Heat absorbed = Mass of ocean water × Specific heat capacity of water × Temperature change
Given that the oceans constitute 97.5% of the Earth's water, the mass of the ocean water is 0.975 times the total mass of water on Earth.
Assuming the oceans have an average temperature of 4.8°C, the temperature change is 4.8°C - 0°C = 4.8°C.
Now we can calculate the change in temperature of the ocean:
Change in temperature = Heat released / (Mass of ocean water × Specific heat capacity of water)
Substituting the values, we get:
Change in temperature = (0.0175 × Total mass of water) × (Specific heat capacity of ice × Temperature change) / (0.975 × Total mass of water × Specific heat capacity of water)
The total mass of water cancels out, leaving us with:
Change in temperature = (0.0175 × Specific heat capacity of ice × Temperature change) / (0.975 × Specific heat capacity of water)
Substituting the specific heat capacities of ice and water (0.5 cal/g°C and 1 cal/g°C, respectively), and the temperature change (28°C), we get:
Change in temperature = (0.0175 × 0.5 cal/g°C × 28°C) / (0.975 × 1 cal/g°C)
Simplifying the equation, we find:
Change in temperature ≈ -0.28°C
Therefore, if all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
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What force acts on a projectile in the horizontal direction?
The force that acts on a projectile in the horizontal direction is Gravitational force.
A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.
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The obliquity of the rotation of Uranus is over 90 degrees. Compared to the plane of the solar system, it rotates on its "side", unlike any other planet. It is surmised that this angle of rotation was caused by:
In Bosnia, the ultimate test of a young man's courage used to be to step off a 400-year-old bridge (destroyed in 1993; rebuilt in 2004) into the River Neretva, 23 m below the bridge. Find a - How long did the drop last?, find b - How fast was the man traveling upon impact with the river?, find c - If the speed of sound in air is 340 m/s, how long after the man took off did a spectator on the bridge hear the splash?.
(a) the drop lasted approximately 2.17 seconds.
(b) the man was traveling at approximately 21.26 m/s upon impact with the river.
(c) approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
(a) To find the time it took for the drop, we can use the equation for free fall motion:
Δy = (1/2) * g * [tex]t^2[/tex]
Given:
Initial height, h = 23 m
Acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]
Rearranging the equation, we get:
t^2 = (2 * h) / g
Substituting the values:
t^2 = (2 * 23 m) / 9.8 [tex]m/s^2[/tex]
t^2 ≈ 4.6949 s^2
Taking the square root of both sides, we find:
t ≈ √(4.6949 [tex]s^2[/tex])
t ≈ 2.17 s
Therefore, the drop lasted approximately 2.17 seconds.
(b) To find the speed of the man upon impact with the river, we can use the equation for final velocity in free fall:
v = g * t
Substituting the values:
v = 9.8 [tex]m/s^2[/tex] * 2.17 s
v ≈ 21.26 m/s
Therefore, the man was traveling at approximately 21.26 m/s upon impact with the river.
(c) To find the time it takes for the sound of the splash to reach a spectator on the bridge, we can use the speed of sound:
Given:
Speed of sound, v_sound = 340 m/s
The time it takes for the sound to travel from the river to the spectator is the same as the time it took for the man to fall. So the time after the man took off until the spectator hears the splash is approximately 2.17 seconds.
Therefore, approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
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Suppose that the mirror is moved so that the tree is between the focus point F and the mirror. What happens to the image of the tree?
A. The image moves behind the curved mirror.
B. The image appears shorter and on the same side of the mirror.
C. The image appears taller and on the same side of the mirror.
D. The image stays the same.
Answer:
C
Explanation:
If the tree is placed between the focus point F and the mirror in a concave mirror, the image of the tree will appear taller and on the same side of the mirror. Therefore, the correct answer is C. The image appears taller and on the same side of the mirror.
Which of the following does not serve as a way to neutralize the charge in a body?
Question 20 options:
A)
Adding more protons to a positively charged body until the number of protons matches the number of electrons
B)
Bringing the charged body into contact with another body having an equal but opposite charge
C)
Adding free electrons to a positively charged body
D)
Allowing free electrons to escape from a negatively charged body
Answer:
A) Adding more protons to a positively charged body until the number of protons matches the number of electrons.
Explanation:
Adding more protons to a positively charged body would only increase the positive charge and further imbalance the charge. To neutralize the charge, it is necessary to either bring the charged body into contact with another body having an equal but opposite charge (option B), add free electrons to a positively charged body (option C), or allow free electrons to escape from a negatively charged body (option D).
A person walks 3.30 km south and then 2.00 km east, all in 3.20 hours. Answer parts a-c.
(a) the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.
(b) Magnitude of average velocity ≈ 3.85 km and Direction of average velocity ≈ -59.04° (south of east)
(c) the average speed during the given time interval is approximately 1.66 km/h.
(a) To find the magnitude and direction of the person's displacement, we can use the Pythagorean theorem and trigonometry.
Displacement in the x-direction = 2.00 km east
Displacement in the y-direction = -3.30 km south (negative because it is in the opposite direction of the positive y-axis)
Using the Pythagorean theorem:
Magnitude of displacement = √((2.00 km)^2 + (-3.30 km)^2)
Magnitude of displacement ≈ 3.85 km
To find the direction, we can use trigonometry:
θ = tan^(-1)(opposite/adjacent)
θ = tan^(-1)(-3.30 km / 2.00 km)
θ ≈ -59.04° (measured counterclockwise from the positive x-axis)
Therefore, the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.
(b) Average velocity is defined as displacement divided by time. The magnitude and direction of average velocity will be the same as the magnitude and direction of displacement.
Magnitude of average velocity ≈ 3.85 km
Direction of average velocity ≈ -59.04° (south of east)
(c) Average speed is defined as total distance traveled divided by time. The total distance traveled is the sum of the magnitudes of the individual displacements.
Total distance = 3.30 km + 2.00 km = 5.30 km
Average speed = Total distance / Time
Average speed ≈ 5.30 km / 3.20 hours
Average speed ≈ 1.66 km/h
Therefore, the average speed during the given time interval is approximately 1.66 km/h.
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7. A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds. What
is the force exerted on the ball by the goalie? (Hint: Apply two formulas to solve this problem)
A. 5 N
B. 10 N
C. 20 N
D. 25 N
A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds then the force exerted on the ball by the goalie is 20N. Option C is correct.
Here, we must determine the change in momentum of the soccer ball. The momentum of an object is stated by the product of its mass and velocity. The mass of the soccer ball is 0.5 kg, and its initial velocity is 10 m/s. Therefore, the ball is conducted to be constant, and its final velocity is 0 m/s.
The change in momentum is computed by reducing the final momentum from the initial momentum. In this concern, the initial momentum is 0.5 kg × 10 m/s = 5 kg·m/s, and the final momentum is 0.5 kg × 0 m/s = 0 kg·m/s. Now, the change in momentum is 5 kg·m/s - 0 kg·m/s = 5 kg·m/s.
Next, we separate the change in momentum by the time taken to bring up the ball to rest, which is 0.25 seconds. Thus, the goalie's force exerted on the ball is 5 kg·m/s / 0.25 s = 20 N.
Therefore, the correct answer is C. 20 N.
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The correct Option is C. The force exerted on the ball by the goalie is 20 N.
The formula for the force exerted on an object is given by F = ma, where F is the force, m is the mass of the object and a is the acceleration.
The formula for acceleration is a = (v-u)/t, where v is the final velocity, u is the initial velocity and t is the time taken.
The acceleration is negative if the object is brought to rest.
So, for the given problem, the initial velocity of the soccer ball is 10 m/s and the final velocity is 0.
The time taken to bring it to rest is 0.25 s.
Therefore, the acceleration is given by:a = [tex](0 - 10)/0.25 = - 40 m/s^{2}[/tex]
Now, we can calculate the force exerted by the goalie using the formula: [tex]F = maF = 0.5 kg $\times$ (- 40 m/s^{2} ) = - 20 N[/tex]
We get a negative value for the force, which means that the force exerted is in the opposite direction to the motion of the ball.
However, the magnitude of the force is given by |-20 N| = 20 N.
So, the answer is option (C) 20 N.
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A model rocket is launched straight upward with an initial speed of 57.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 140 m. Answer parts b-d.
a. The maximum height reached by the rocket is 1083 meters.
b. The rocket reaches its maximum height 38 seconds after liftoff.
c. The rocket is in the air for 1.09 seconds.
How do we calculate?(b)
We will apply equation of motion :
v² = u² + 2aΔy
Δy = (v² - u²) / (2a)
Δy = (0 - 57.0²) / (2 * 1.50)
Δy = (-57.0)² / 3.00
Δy = 3,249 / 3.00
Δy = 1083 m
(c)
v = u + at
0 = u + at
t = -u / a
t = -57.0 / 1.50
t = 38 seconds
(d)
Δy = ut + (1/2)at²
140 = 57.0t + (1/2)(1.50)t²
(1/2)(1.50)t² + 57.0t - 140 = 0
t = 1.09 seconds.
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Masses m and 2m are joined by a light inextensible string which runs without slipping over a uniform circular pulley of mass 2m and radius a. Using the angular position of the pulley as generalized coordinate, write down the Lagrangian function and Lagrange's equation. Find the acceleration of the masses.
Answer: the acceleration of the masses is given by = 0, which means the angular acceleration of the pulley is zero. This implies that the masses m and 2m move with constant velocity, they are in equilibrium.
Look at the velocity versus time graph below. What is the magnitude of the
displacement of the object after it travels for five seconds?
Velocity (m/s)
Time (s)
A. 30 m
OB. 20 m
OC. 25 m
OD. 35 m
The magnitude of displacement of the object after five seconds, calculated from the velocity-time graph, is 32.5 m. The correct answer is option E.
Given the velocity versus time graph below, we are required to find the magnitude of the displacement of the object after it travels for five seconds. Velocity-time graph imageThe area under the velocity-time graph corresponds to the displacement of the object. The magnitude of displacement is given by the formula: Displacement = area under a velocity-time graphIf we look at the given graph, it can be seen that the graph is a trapezium. Therefore, we need to split it into two parts: a rectangle and a triangle. The displacement is given by the sum of the area of both parts. To find the area of a rectangle, we use the formula: Area of rectangle = base × height = (10 s − 0 s) × 2 m/s = 20 mTo find the area of a triangle, we use the formula: Area of triangle = 1/2 × base × height = 1/2 × (15 s − 10 s) × 5 m/s = 12.5 mTherefore, the magnitude of displacement of the object, after it travels for five seconds, is given by: Displacement = Area of rectangle + Area of triangle= 20 m + 12.5 m= 32.5 mHence, the correct answer is option E. 32.5 m.For more questions on the magnitude of displacement
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An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 44.5 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?
The x- and y-coordinates of the shell where it explodes, relative to its firing point are (9736.5 m, 762.3 m) respectively.
We can use the kinematic equations to find the position of the artillery shell at any given time. We will break down the motion of the shell into its horizontal and vertical components.
First, we can find the initial horizontal and vertical velocities of the shell as follows:
\begin{align} v_{0x} &= v_0 \cos(\theta) = 300 \cos(52.0^\circ) \approx 192.9\text{ m/s}\ v_{0y} &= v_0 \sin(\theta) = 300 \sin(52.0^\circ) \approx 245.4\text{ m/s} \end{align}
We can use the vertical motion of the shell to find the time it takes to reach its maximum height, using the following kinematic equation:
$$y = v_{0y}t - \frac{1}{2}gt^2$$
At maximum height, the vertical velocity will be zero, so we can solve for the time it takes to reach this point:
\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y}}{g} \approx 25.2\text{ s} \end{align}
Therefore, the time it takes for the shell to reach maximum height is 25.2 seconds. Using this time, we can find the maximum height, as follows:
\begin{align} y_\text{max} &= v_{0y}t - \frac{1}{2}gt^2\ &= 245.4\text{ m/s} \cdot 25.2\text{ s} - \frac{1}{2}(9.81\text{ m/s}^2)(25.2\text{ s})^2\ &\approx 762.3\text{ m} \end{align}
The time it takes for the shell to hit the mountainside can be found by solving for the time when y = 0:
\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y} + \sqrt{(v_{0y})^2 + 2gy_\text{max}}}{g} \approx 50.5\text{ s} \end{align}
Therefore, the time it takes for the shell to hit the mountainside is 50.5 seconds. The x-coordinate of the explosion can be found by using the horizontal velocity and the time it takes for the shell to hit the mountainside:
\begin{align} x &= v_{0x}t\ &= 192.9\text{ m/s} \cdot 50.5\text{ s}\ &\approx 9736.5\text{ m} \end{align}
Therefore, the x-coordinate of the explosion is 9736.5 meters. The y-coordinate of the explosion is simply the height of the mountainside:
$$y = 0 + 762.3\text{ m} = 762.3\text{ m}$$
Therefore, the y-coordinate of the explosion is 762.3 meters.
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What are the six digit grid coordinates for the windtee?
The six digit grid coordinates for the windtee should be 3.
How do we we calculate?The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.
When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.
If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.
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The six digit grid coordinates for the windtee is determined as 100049.
What is a coordinate point?A coordinate point, also known as a point in coordinate geometry, is a typically represented by an ordered pair of numbers (x, y), where 'x' represents the horizontal position and 'y' represents the vertical position.
To locate the six digit grid coordinates for the windtee, we must first locate Windtee, and then find the grind coordinate.
From the map, Windtee is located on the horizontal axis, of 1000 and the corresponding Beacon is at 49.
So the six digit grid coordinates = 100049.
Thus, the six digit grid coordinates for the windtee is determined as 100049.
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what is a shargaff rule
According to Chargaff's rule, the amounts of adenine (A), thymine (T), and guanine (G) in the DNA molecule are equal to each other. The amounts of cytosine (C) and guanine (G) are also equal.
Who is Chargaff ?Erwin Chargaff was a biochemist, author, Bucovinian Jew who immigrated to America during the Nazi era, and professor of biochemistry at Columbia University's medical school.
Chargaff found patterns among the four bases, or chemical building blocks, of DNA, which are directly related to DNA's function as the genetic material of living things.
He was born in Austria-Hungary. Heraclitean Fire: Sketches from a Life Before Nature, an autobiography he penned, received positive reviews.
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A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 13.5 m/s, and the man is initially 3.55 m above the level of the saddle. Find a - What must be the horizontal distance between the saddle and limb when the man makes his move? Find b - How long is he in the air?
(a) the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.
(b) the man is in the air for approximately 0.843 seconds.
To determine the horizontal distance between the saddle and limb when the man makes his move, we need to consider the horizontal velocity of the man when he drops from the tree limb.
Given:
Speed of the horse (constant velocity), v = 13.5 m/s
Vertical distance between the limb and saddle, h = 3.55 m
a) To find the horizontal distance, we can use the formula:
horizontal distance = horizontal velocity × time
Since the man drops vertically, his initial horizontal velocity is zero. The only horizontal velocity he will have is due to the motion of the horse.
The time taken by the man to fall can be determined using the equation for free fall:
h = (1/2) × g × t²
Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.
Rearranging the equation, we get:
t = √(2h / g)
Substituting the given values:
t = √(2 × 3.55 / 9.8) ≈ 0.843 s
Now, we can find the horizontal distance:
horizontal distance = v × t
horizontal distance = 13.5 × 0.843 ≈ 11.386 m
Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.
b) The time the man is in the air can be calculated using the same equation for free fall:
t = √(2h / g)
Substituting the given value of h:
t = √(2 × 3.55 / 9.8) ≈ 0.843 s
Thus, the man is in the air for approximately 0.843 seconds.
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Two particles move about each other in circular orbits under the influence of gravitational forces, with a period 7, Their motion is suddenly stopped at a given instant of time and they are then released and allowed to fall into each other......
Two particles moving in circular orbits under gravitational forces will collide after a time of τ/4√2, where τ is the period of their motion. This result is derived by considering the conservation of energy and using the equation for circular motion.
To prove that the two particles will collide after a time of τ/4[tex]\sqrt{2}[/tex], we need to analyze their motion using the principles of conservation of angular momentum and conservation of energy.
Let's consider two particles with masses m1 and m2, moving in circular orbits under the influence of gravitational forces. The period of their motion is given as τ.
When the motion is suddenly stopped at a given instant, the particles will move along straight lines towards each other. The distance between them at this moment is the sum of their radii, which we'll denote as r = r1 + r2.
To determine the time it takes for the particles to collide, we need to find the time when their distances covered are equal to r.
Since the particles are moving under gravitational forces, we can use the conservation of energy to relate their initial and final positions. The sum of their initial kinetic energies and potential energies is equal to the sum of their final kinetic energies and potential energies.
Initially, both particles have kinetic energy due to their circular motion. When the motion is stopped, their kinetic energies become zero. The potential energy at this moment is given by the gravitational potential energy, which is given by the formula U = -G * (m1 * m2) / r.
Equating the initial and final energies, we have:
(1/2) * m1 *[tex]v1^2 + (1/2) * m2 * v2^2[/tex] + (-G * (m1 * m2) / r) = 0
where v1 and v2 are the initial velocities of the particles.
Since the particles start from rest, their initial velocities are zero.
Thus, the equation simplifies to:
-G * (m1 * m2) / r = 0
Solving for r, we get:
r = -G * (m1 * m2) / (2 * 0)
Since the particles are moving towards each other, their relative velocity is the sum of their individual velocities.
[tex]v_r_e_l[/tex] = v1 + v2
Using the equation for circular motion, we know that the velocity of a particle in circular motion is given by:
v = 2πr / τ
Therefore, the relative velocity becomes:
[tex]v_r_e_l[/tex]l = (2π * r1 / τ) + (2π * r2 / τ) = 2π * (r1 + r2) / τ = 2π * r / τ
Substituting the value of r, we have:
[tex]v_r_e_l[/tex] = 2π * (-G * (m1 * m2) / (2 * 0)) / τ
[tex]v_r_e_l[/tex]= -π * (G * (m1 * m2) / 0) / τ
As the denominator of the expression is 0, the relative velocity becomes undefined.
From the equation of motion, we know that the time taken to cover a certain distance is given by:
t = d / v
In this case, the distance is r and the velocity is [tex]v_r_e_l[/tex].
Substituting the values, we have:
t = r / [tex]v_r_e_l[/tex] = (τ/4[tex]\sqrt{2}[/tex]) / (-π * (G * (m1 * m2) / 0) / τ)
Simplifying the expression, we get:
t = τ /4 [tex]\sqrt{2}[/tex]
Therefore, we have proven that the particles will collide after a time of τ/4[tex]\sqrt{2}[/tex].
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A projectile is launched with an initial speed of 48.0 m/s at an angle of 34.0° above the horizontal. The projectile lands on a hillside 3.65 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) Answer parts a-b.
(a) The projectile's velocity at the highest point of its trajectory is approximately 27.01 m/s, counterclockwise from the +x-axis.
(b) The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.
To solve this problem, we'll analyze the projectile's motion in two dimensions: horizontal and vertical.
(a) To find the projectile's velocity at the highest point of its trajectory, we need to consider the vertical component and horizontal component separately.
The initial velocity (V0) of the projectile is 48.0 m/s, and the launch angle (θ) is 34.0° above the horizontal.
The vertical component of velocity (Vy) can be found using the equation:
Vy = V0 * sin(θ)
Plugging in the known values:
Vy = 48.0 m/s * sin(34.0°)
Calculating Vy, we find:
Vy ≈ 27.01 m/s
The horizontal component of velocity (Vx) can be found using the equation:
Vx = V0 * cos(θ)
Plugging in the known values:
Vx = 48.0 m/s * cos(34.0°)
Calculating Vx, we find:
Vx ≈ 39.79 m/s
Therefore, at the highest point of its trajectory:
- The magnitude of the projectile's velocity is approximately 27.01 m/s.
- The direction of the velocity is straight up, counterclockwise from the +x-axis.
(b) To find the straight-line distance from where the projectile was launched to where it hits its target, we need to consider the horizontal motion of the projectile.
The time of flight (t) is given as 3.65 s.
The horizontal distance (x) can be found using the equation:
x = Vx * t
Plugging in the known values:
x = 39.79 m/s * 3.65 s
Calculating x, we find:
x ≈ 144.93 m
The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.
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When a piece of wood is put in a graduated cylinder containing water the level of water rises from 17.7cm cubic to 18.5cm cubic calculate the total volume of the piece of wood given that it's relative density is 0.60
The total volume of the piece of wood is 1.33[tex]cm^3[/tex].
To calculate the total volume of the piece of wood, we can use the principle of displacement.
1. First, we need to find the difference in volume between the two water levels. The initial volume is 17.7 [tex]cm^3[/tex], and the final volume is 18.5 cm^3. The difference is 18.5 [tex]cm^3[/tex] - 17.7 [tex]cm^3[/tex] = 0.8 [tex]cm^3[/tex].
2. Now, we need to find the volume of water displaced by the piece of wood. Since the relative density of the wood is 0.60, it means that the wood is 0.60 times denser than water.
3. The volume of water displaced by the wood is equal to the difference in volume divided by the relative density of the wood. So, the volume of water displaced is 0.8 cm^3 / 0.60 = 1.33 [tex]cm^3[/tex].
4. Finally, the total volume of the piece of wood is equal to the volume of water displaced. Therefore, the total volume of the piece of wood is 1.33 [tex]cm^3[/tex].
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WHOEVER ANSWERS IS THE BRAINLIEST!!! PLS HELP!!
Based on the information, we can infer that the temperature on the west and east coasts of the United States is higher than in the central part at latitude 35° North.
What do we see in the image?In the image you can see the map of the United States and two latitudinal lines of 35° and 45° North. Additionally we see the different temperatures that exist in various cities or locations in the United States.
Based on this information, we can infer that the temperatures on the east and west coasts are higher than the temperatures recorded in the central part. For example, at 35° latitude, the coasts register temperatures of more than 60°F while the central zone registers lower temperatures between 36 and 59°F.
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A woman stands at the edge of a cliff and throws a pebble horizontally over the edge with a speed of v0 = 20.5 m/s. The pebble leaves her hand at a height of h = 55.0 m
above level ground at the bottom of the cliff, as shown in the figure. Note the coordinate system in the figure, where the origin is at the bottom of the cliff, directly below where the pebble leaves the hand. Answer parts a-f.
(a)The time taken for the pebble to reach the ground is approximately 2.01 seconds, and
(b) the horizontal distance traveled by the pebble is approximately 41.02 meters.
(c) The vertical distance traveled by the pebble is 55 meters.
(d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.
(e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.
(f) The negative sign indicates that the pebble is moving downward.
a) To find the time taken for the pebble to reach the ground, we can use the equation for vertical motion:
h = (1/2)gt^2, where h is the vertical distance and g is the acceleration due to gravity.
Rearranging the equation, we have:
t = √((2h) / g), where t is the time taken.
Substituting the given values, we get:
t = √((2 * 55) / 9.8) ≈ 2.01 seconds.
b) The horizontal speed of the pebble remains constant throughout its motion. Therefore, the horizontal distance traveled by the pebble can be found by multiplying the horizontal speed by the time taken:
d = v0 * t, where d is the horizontal distance and v0 is the initial horizontal speed.
Substituting the given values, we have:
d = 20.5 * 2.01 ≈ 41.02 meters.
c) The vertical distance traveled by the pebble is given as 55 meters.
d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.
e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.
f) The final vertical velocity of the pebble when it reaches the ground can be found using the equation:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the initial vertical velocity is 0 m/s and the acceleration due to gravity is -9.8 m/s^2, we have:
v = 0 + (-9.8) * 2.01 ≈ -19.8 m/s.
The negative sign indicates that the pebble is moving downward.
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11. The figure shows a block of mass M = 7.75 kg hanging at rest. The light wire fastened to the wall is
horizontal and has a tension of 38 N. The wire fastened to the ceiling is also very light, has a tension
of 59 N and makes an angle with the ceiling. Find the angle 8.
QUA
The angle made with the ceiling by the tension force of the two wires is determined as 50⁰.
What is the angle made by the two tensions?The angle made by the two tensions is calculated by applying cosine rule as follows;
the force opposite the angle = weight of the block = mg
W = 7.75 kg x 9.8 m/s²
W = 75.95 N
The angle made by the two tensions is calculated as follows;
cos θ = (38 N ) / ( 59 N)
cos θ = 0.6441
θ = arc cos (0.6441)
θ = 50⁰
Thus, the angle made by the two tensions is determined as 50 degrees.
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The figure is in the image attached
a current of a 6 flows through a light bulb for 12 s, how many coulombs of charge pass through the light bulb during this time
A current of a 6 flows through a light bulb for 12 s. The total charge that passes through the light bulb during the given time is 72 coulombs.
To calculate the total charge that passes through the light bulb, we need to use the formula Q = I * t, where Q represents the charge in coulombs, I represents the current in amperes, and t represents the time in seconds.
Step 1: Identify the known values:
Current (I) = 6 amperes
Time (t) = 12 seconds
Step 2: Calculate the charge using the formula:
Q = I * t
Step 3: Substitute the known values into the formula:
Q = 6 amperes * 12 seconds
Q = 72 coulombs
Therefore, the total charge that passes through the light bulb during the given time is 72 coulombs.
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A 17-kg
piece of metal displaces 2.8 L
of water when submerged. what is its density?
Answer: Density = 6071.428571 kg/m³
Explanation: Given that mass m=17 kg
volume displaced v=2.8L
We know that
density = mass/volume
Here density=17kg/2.8L
Also 1L=1000m³ Hence
density=17kg/2.8×10⁻³m³
=6071.428571 kg/m³
2. Explain brightness of light using the wave model of light
The brightness of light is explained by the wave model of light. Brightness refers to the perceived intensity of light. Brightness is determined by the amplitude or intensity of light waves.
The larger the amplitude, the brighter the light. This can be explained using the wave model of light.Light is a form of electromagnetic radiation that is composed of oscillating electric and magnetic fields. The wave model of light states that light is a transverse wave that propagates through space. The wave model of light states that light travels in straight lines and can be reflected, refracted, and diffracted. Brightness is a measure of the intensity of light waves.
The intensity of light waves is determined by the amplitude of the wave.The amplitude of a wave is the maximum displacement of the wave from its equilibrium position. The larger the amplitude of a wave, the more energy the wave carries. This means that the larger the amplitude of light waves, the brighter the light. The brightness of light can be increased by increasing the amplitude of light waves. This can be achieved by increasing the intensity of light waves. The intensity of light waves can be increased by increasing the power of the light source. Thus, brightness can be explained by the wave model of light as it is determined by the amplitude or intensity of light waves.
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Select the correct answer.
Before a collision, the x-momentum of an object is 8.0 × 103 kilogram meters/second, and its y-momentum is 1.2 × 104 kilogram meters/second. What is the magnitude of its total momentum after the collision?
A.
1.4 × 104 kilogram meters/second
B.
2.0 × 104 kilogram meters/second
C.
3.2 × 104 kilogram meters/second
D.
5.7 × 104 kilogram meters/second
PLEASE HELP ALL I NEED IS A DRAWING, i posted this like 100 times please help.
Answer:
Hope this helps
Explanation:
deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W. Refer to appropriate Newton’s laws of motion.
The expression for the contribution of P to the pressure exerted on W is P = mV/(c^2t), derived using Newton's laws of motion and the definition of pressure.
In order to deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W, we can use the appropriate Newton’s laws of motion. Specifically, we can use the equation F = ma, where F represents force, m represents mass, and a represents acceleration.We know that pressure (P) is defined as force per unit area, or P = F/A. Rearranging this equation, we can solve for force: F = PA.Substituting this into the equation F = ma, we get PA = ma. Rearranging this equation, we can solve for pressure in terms of mass and acceleration: P = ma/A. Finally, we know that acceleration can be expressed in terms of velocity (V) and time (t): a = V/t.Substituting this into our equation for pressure, we get P = mV/(At). Since c represents the speed of sound, we can express A as [tex]A = c^2[/tex]. Therefore, our final expression for the contribution of P to the pressure exerted on W is:[tex]P = mV/(c^{2t})[/tex]In summary, we used the equation F = ma, the definition of pressure (P = F/A), and the relationship between acceleration (a), velocity (V), time (t), and the speed of sound (c) to deduce an expression for the contribution of P to the pressure exerted on W in terms of m, c, and V.For more questions on pressure
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Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.0 kg) portrayed in the figure below move on a frictionless surface and a force F = 34 N acts as shown on the 4.0-kg block. Answer parts a-c.
(a) The acceleration of the system is 8.5 m/s².
(b) The tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.
(c) The force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.
To solve this problem, we can use Newton's second law of motion (F = ma) and consider the forces acting on each block individually.
(a) Determine the acceleration given this system:
To find the acceleration (a) of the system, we can use the net force acting on the 4.0 kg block (m3). The only force acting on m3 is the applied force (F = 34 N).
F = m3 * a
34 N = 4.0 kg * a
Solving for a, we find:
a = 34 N / 4.0 kg
a = 8.5 m/s²
Therefore, the acceleration of the system is 8.5 m/s².
(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks:
To find the tension in the cord (T), we can consider the forces acting on the 1.0 kg block (m1).
T - F = m1 * a
T - 34 N = 1.0 kg * 8.5 m/s²
T - 34 N = 8.5 N
T = 42.5 N
Therefore, the tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.
(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block:
To find the force exerted by the 1.0 kg block (m1) on the 2.0 kg block (m2), we can consider the forces acting on the 2.0 kg block.
F - T = m2 * a
F - 42.5 N = 2.0 kg * 8.5 m/s²
F - 42.5 N = 17 N
F = 59.5 N
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.
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