The two types of potential energies which would be types of mechanical energies are elastic and gravitational.
The correct answer choice is option D
What is meant by potential energy?Potential energy means the energy a body possess in relation to its position. Other forms of energy include
Electrical energyKinetic energyChemical energySound energyNuclear energy Thermal energy Radiant energyKinetic energy is the energy a body possesses due to its motion.
So therefore, the two types of potential energies which would be types of mechanical energies are elastic and gravitational. The correct answer choice is option D
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Answer: it is d
Explanation: i took the unit test on edge
Quick Quiz 40.2 While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these types of waves in terms of photon energy from highest to lowest
The order of decreasing photon energy is FM radio, AM radio, yellow light, and microwaves.
Electromagnetic radiation: The electromagnetic field's waves, which are conveying electromagnetic radiant energy through space, are what make up electromagnetic radiation. It consists of X-rays, gamma rays, microwaves, infrared, light, and radio waves. These waves are all a component of the electromagnetic spectrum.
Microwave radiation has wavelengths between one meter and one millimeter, which correspond to frequencies between 300 MHz and 300 GHz, respectively.
Radio waves: The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz and below.
The radio waves have the highest photon energy and the lowest is microwaves.
So, the highest to lowest order is as follows:
FM radio, AM radio, yellow light, and microwaves.
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A reconnaissance plane flies 430 km away from its base at 680 m/s, then flies back to its base at 1020 m/s. what is its average speed? answer in units of m/s.
The average speed of the plane is 816 m/s.
We need to know about the average speed to solve this problem. The average speed can be determined as the total distance divided by the total time. It can be written as
v = d / t
where v is average speed, d is distance and t is time.
From the question above, we know that
d1 = d2 = 430 km
v1 = 680 m/s
v2 = 1020 m/s
Because of the distance is the same, we can calculate the average speed as
v = 2d / (t1 + t2)
v = 2.d / (d/v1 + d/v2)
v = 2/(1/v1 + 1/v2)
v = 2/(1/680 + 1/1020)
v = 816 m/s
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A uniform magnetic field of magnitude 0.150 T is directed along the positive x axis. A positron moving at a speed of 5.00×10⁶ m/s enters the field along a direction that makes an angle of Ф = 85.0° with the x axis (Fig. P 29.73 ). The motion of the particle is expected to be a helix as described in Section 29.2. Calculate (a) the pitch p
The pitch of the helical path of positron is, [tex]p=1.038 \times 10^{-4}[/tex]
Explain the trajectory's radius?The radius of a circular arc that most closely resembles the curve at a given point on a projectile's trajectory is the radius of the trajectory, which reveals the direction in which the projectile is curving at that particular location.
Given:
Magnitude of magnetic field, [tex]B=0.150\;T[/tex]
Speed of position, [tex]v=5 \times 10^{6}\;m/s[/tex]
Angle, [tex]\;\theta=85^{\circ}[/tex]
The motion of the particle is expected to be a helix.
(a)
We know that,
[tex]Pitch,\;p=V_{11} \times T[/tex]
Where,
[tex]V_{11}=parallel\;velocity\;to\;magnetic\;field[/tex]
[tex]T=Time\;period=\frac{2 \times \pi \times m}{q \times B}[/tex]
[tex]Mass\;of\;positron,\;m=9.1 \times 10^{-31}\;kg\\[/tex]
[tex]Change\;of\;positron,\;q=+1e=1.6 \times 10^{-19}\;kg[/tex]
Substitute the known values in pitch equation,
[tex]p=Vcos\theta\frac{2\times \pi \times m}{q\times B}[/tex]
[tex]p=(5\times 10^{6)} cos85^{\circ}\;\frac{2\times \pi \times 9.1 \times 10^{-31} }{1.6 \times 10^{-19} \times 0.150}[/tex]
By solving the above equation, we get
[tex]p=1.038 \times 10^{-4}[/tex]
Therefore,
[tex]p=1.038 \times 10^{-4}[/tex]
(b)
We know that,
[tex]Radius\;of\;trajectory,\;r=\frac{m \times V_{1} }{q \times B}[/tex]
Where,
[tex]V_{1}=perpendicular\;velocity\;to\;magnetic\;field[/tex]
[tex]m=mass\;of\;positron=9.1 \times 10^{-31} \;kg[/tex]
[tex]q=charge\;of\;positron=+1e=1.6 \times 10^{-19}\;kg[/tex]
Substitute the known values in the radius of trajectory equation,
[tex]Radius\;of\;trajectory,\;r=\frac{m}{q \times B}\;(Vsin\theta)[/tex]
[tex]Radius\;of\;trajectory,\;r=\frac{9.1 \times 10^{-31} \times 5 \times 10^{6} \times sin85^{\circ}}{1.6 \times 10^{-19} \times 0.150}[/tex]
By solving the above equation, We get,
[tex]r=1.88 \times 10^{-4} \;m[/tex]
Therefore, The radius of the trajectory is,
[tex]r=1.88 \times 10^{-4} \;m[/tex]
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buoyancy: a steel ball sinks in water but floats in a pool of mercury, which is much denser than water. where is the buoyant force on the ball greater?
The buoyant force on the ball in the pool of mercury is greater.
What do you mean by buoyant force?An object is subject to the buoyant force of the liquid ( fully or partially submerged in a liquid ). An object won't sink in a liquid if its buoyant force (pointing upward) exceeds its weight (pointing downward) on the object. A fluid's buoyancy, also known as its upthrust, is the upward force it exerts in opposition to the weight of an object that is partially or completely submerged. The weight of the fluid on top causes pressure in a fluid column to rise with depth. As a result, the pressure at the bottom of a fluid column is higher than at the top.
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HELP!!
A football is kicked straight up into the air; it hits the ground
5.2 s later.
a) What was the greatest height reached by the ball? Assume l
is kicked from ground level.
b) With what speed did it leave the kicker's foot?
what is the major cause of the earth’s magnetic field? the earth’s liquid outer core the earth’s liquid outer core the earth’s magnetic lithosphere the earth’s magnetic lithosphere charged particles from the sun charged particles from the sun magnetic materials on the earth’s surface
The earth's liquid outer core is the major cause of the earth’s magnetic field.
What is magnetic field?The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, a vector field. A force acting on a charge while it travels through a magnetic field is perpendicular to both the charge's motion and the magnetic field. The magnetic field of a permanent magnet attracts or repels other magnets as well as ferromagnetic elements like iron. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilised in electromagnets, and electric fields that change over time produce magnetic fields that surround magnetised things.
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Suppose our Sun is about to explode. In an effort to escape, we depart in a spacecraft at v=0.800 c and head toward the star Tau Ceti, 12.0 ly away. When we reach the midpoint of our journey from the Earth, we see our Sun explode, and, unfortunately, at the same instant, we see Tau Ceti explode as well.(a) In the spacecraft's frame of reference, should we conclude that the two explosions occurred simultaneously? If not, which occurred first?
Sun exploded 16 years after Tau Ceti.
Considering a scenario in which the Sun is going to explode.
In order to escape the explosion, we depart in a spacecraft with a speed of v = 0.8c.
The star Tau Ceti is 12 life years away.
At the midpoint of the journey, the Sun as well as Tau Ceti explode. at the same instant.
Now, the distance between the Sun and the spacecraft:
L = L(p) √( 1 - (v/c)²)
L = (6 ly) √[1 - (0.8)²] = 3.6 ly
Now, we observe Sun to be flying away from us at a speed of 0.8c whereas the light from the Sun approaches at a speed 1c. Therefore, the gap between the Sun and the blast of the Sun wave is opened at a speed of 1.8c.
Therefore, the time passed ever since the Sun exploded is:
t = 3.60 ly / 1.80c = 2 years.
Now, we observe Tau Ceti moving toward us at a speed of 0.8c, whereas the light from the blast approaches at a speed of 1c which is only 0.2c faster. Then the gap between the Tau Ceti and its blast wave is 3.60 ly and growing at a speed of 0.2c.
Therefore, the time for which it has been opening is:
t = 3.60 ly/ 0.20c
t = 18 years.
Therefore, Tau Ceti exploded 16 years before the Sun.
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When alcohol is rubbed on your body, it lowers your skin temperature. Explain this effect.
When alcohol is rubbed on your body, it lowers your skin temperature due to evaporative cooling.
What is Boiling point?The temperature at which the vapor pressure of a liquid is equal to the pressure of the atmosphere on the liquid and varies according to substances and their components.
When alcohol is rubbed on your body, it lowers your skin temperature, it evaporates and is referred to an endothermic reaction which entails absorption by the body.
Alcohol has a very low boiling point which is why rubbing it on the body will lead to evaporative cooling which will produce a cooling effect and is the most appropriate choice.
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A uniform, rectangular block made of steel has dimensions 1. 00 cm×2. 00 cm×3. 00 cm. It is placed on a wooden board, and then the board is tilted until the block begins to slide down the board. Which face of the block should be in contact with the board, so that it begins to slide at the shallowest angle?.
Dimensions of a steel block that is uniformly rectangular are 1. 00 cm, 2. 00 cm, and 3. 00 cm. It starts to slide at the shallowest angle whenever the block's face comes into contact with the board.
What do you mean by an angle?An angle is formed when the two straight lines or rays meet at a common endpoint. The common point of the contact is called the vertex of an angle. The word angle comes from Latin word named ‘angulus,’ meaning “corner.”
In general, an angle is a figure created in Euclidean geometry by two rays that have a common terminus and are referred to as the vertices and sides of the angle, respectively.
The angles produced by the beams are always contained in the plane that comprises two rays.
In conclusion, the block will start to slide at the angle that offers the least resistance once its face makes contact with the board.
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A wire 2.80m in length carries a current of 5.00A in a region where a uniform magnetic field has a magnitude of 0.390T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (c) 120° .
The force on the wire will be
F = 3.16 Newton
We have a current carrying wire in a uniform magnetic field.
We have to determine the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is 60.0°.
What is the magnitude of force acting on a current (I) carrying wire of length (L) placed in a Magnetic field (B)?The force on the current carrying wire will be -
F = IBL sinθ
According to the question, we have -
L = 2.80 m
I = 5 A
B = 0.39 T
Therefore -
F = IBL sinθ
F = 5 x 0.39 x 2.80 x sin(120)
F = 5.46 x sin(120)
F = 5.46 x 0.58
F = 3.16 Newton
Hence, the force on the wire will be
F = 3.16 Newton
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The magnitude of the magnetic force on the wire 4.728 N.
To find the magnetic force, the values given are:
Length A = 2.80m
Current I = 5 A
Magnetic field B = 0.390 T
Angle = 120° .
What is meant by magnetic force?
The attraction or repulsion that arises between electrically charged particles because of their motion is called Magnetic force.Magnetic force is the basic force which is responsible for such effects as the action of electric motors and the attraction of magnets.A force experienced through the magnetic field with some moving charges is perpendicular to its own velocity.The formula to find the magnetic force,F =iBLsinθ Newton
Substituting the values given,
Magnetic force,
F = 5 A × 0.390 T × 2.80 m × Sin 120°
= 5 A × 0.390 T × 2.80 m × 0.8660254
F = 4.728 N.
The magnitude of the magnetic force on the wire 4.728 N.
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A car travels 10km for 3h, then travels 15km for another 6h, finally travels 3km for 5h. What is the average speed or the car?
The average speed of the car that travels 10km for 3h, then travels 15km for another 6h, finally travels 3km for 5h is 2km/hr.
How to calculate average speed?The average speed of a moving object can be calculated by dividing the distance moved by the substance by the time taken.
Average speed = Distance/time
According to this question, a car travels 10km for 3h, then travels 15km for another 6h, finally travels 3km for 5h.
The total distance of the car is 10km + 15km + 3km = 28kmThe total time = 3h + 6h + 5h = 14hAverage speed = 28km ÷ 14h = 2km/hr
Therefore, the average speed of the car that travels 10km for 3h, then travels 15km for another 6h, finally travels 3km for 5h is 2km/hr.
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A 25 kg block is pulled with an applied force of 200 N across a horizontal surface. The block experiences a frictional force of 75 N.a. Draw a free-body diagram for the block, including all forces acting on the block.
b. Determine the net force acting on the block.
c. Calculate the acceleration of the block.
Answer:
see attachment for the answer.
Explanation:
Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (c) the magnitude and direction of the current in wire 3 .
The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
The force per unit length between two parallel thin current-carrying [tex]I_1[/tex] and [tex]I_2[/tex] wires at distance ' r ' is given by [tex]f=\frac{u_0I_1I_2}{2\pi r}[/tex] ....(1) . If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3
[tex]F_2_1=F_2_3[/tex]
Using equation (1) , we get
[tex]\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A[/tex]
I₃ = 2.4 A and the current is pointing in the downward direction
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The complete question is given below :
A schematic of the information provided in the question can be seen in the image attached below.Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (c) the magnitude and direction of the current in wire 3 .
Which of the following is most likely to form when tectonic plates move away from each other?
Group of answer choices Below.
A. Continental Rift
B. Fault
C. Subduction Zone
D. Trenches
Answer:
Continental Rift
Explanation:
I had earth science in 2020-2021 my freshman year. Rifting can be caused when hot material from a mantle plume reaches the base of a continental plate and causes the overlying lithosphere to heat up. In addition to this the upwards movement of the plume against the base of the plate results in extensional forces which can cause rifting.
A proton is projected in the positive x direction into a region of a uniform electric field →E =(-6.00 × 10⁵) i^ N/C at t=0 . The proton travels 7.00 cm as it comes to rest. Determine (b) its initial speed, and
The initial speed of the proton that is projected in the positive x direction into a region of a uniform electric field is 2.84×〖10〗^6 m/s
The equation of motions expresses the speed of an object with initial velocity u, final velocity v, and distance s, as;
v^2= u^2+2 a.s
Making u the subject of the formula,
u^2= -2 a.s
Given that:
a=5.76×〖10〗^13 m/s^2
s =7.00cm or 0.07m
Inserting values,
u^2= -2 ×(-5.76×〖10〗^13 m/s^2)×0.07m
u^2=8.06×〖10〗^12 m^2/s^2
Therefore,
u= √8.06×〖10〗^12 m^2/s^2
u=2.84×〖10〗^6 m/s
Therefore, the initial speed of the proton is 2.84×〖10〗^6 m/s
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The rotor in a certain electric motor is a flat, rectangular coil with 80 turns of wire and dimensions 2.50 cm by 4.00 cm . The rotor rotates in a uniform magnetic field of 0.800 T . When the plane of the rotor is perpendicular to the direction of the magnetic field, the rotor carries a current of 10.0 mA . In this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn steadily at an angular speed of 3.60×10³ rev/min. (b) Find the peak power output of the motor.
The Peak power output of the motor is 0.452 Watt
To find the peak power output, the given values are,
No .of turns of the wire = 80
Dimensions is given as, 2.50 cm by 4.00 cm
Magnetic field = 0.800 T.
current = 10 mA.
Angular speed = 3.60 ×10³ rev/min
What is Peak power output?Peak power output can be also known as Peak work rate.If the output was greatest or the work production was very high in the given amount of time then it is known as Peak power output.This peak power is depends on the force, distance and time.The formula for peak power output,
Pout = τ * ω
where,
Pout - output power watts (W),
τ - torque Newton meters (Nm),
ω - angular speed radians per second (rad/s).
Calculating angular speed as here rotational speed of the motor in rpm is given:
ω = rpm * 2π / 60
where,
ω – angular speed, radians per second (rad/s),
rpm – rotational speed in revolutions per minute,
π – mathematical constant pi (3.14),
60 – number of seconds in a minute.
So, the formula is,
Peak power output = τ *rpm * 2π / 60
τ = NIAB sinθ Nm
= 80×10×10⁻³×0.0250×0.040×0.800× Sin 90°
= 2 × 6.4 × 10 ⁻⁴
= 1.2 × 10⁻⁴ Nm.
Peak power output = 1.2 × 10⁻⁴ ×3.60×10³ × 2π / 60
= 0.452 Watt
Thus, the peak power output is 0.452 Watt.
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How do I solve this???
If a toy rocket is launched vertically and rises to a height of 250 meters, then the initial velocity of the toy rocket would be 70 m / s.
What are the three equations of motion?
There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
As given in the problem, a toy rocket is launched vertically and rises to a height of 250 meters, we have to find the initial velocity of the launch,
By using the third equation of motion given by Newton,
v² - u² = 2×a×s
0² - u² = 2×(-9.81)×250
u² = 4905
u = 70 m / s
Thus, the initial velocity of the toy rocket would be 70 m / s.
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A motorboat cuts its engine when its speed is 10. 0m/s and then coasts to rest. The equation describing the motion of the motorboat during this period is v=vi e^⁻ct, where v is the speed at time t, vi is the initial speed at t=0 , and c is a constant. At t=20. 0s , the speed is 5. 00m/s (c) Differentiate the expression for v(t) and thus show that the acceleration of the boat is proportional to the speed at any times
The acceleration of the motorboat is proportional to the speed.
The speed of the motorboat when it cuts its engine =
10 m/s
The motion of a motorboat is v.
[tex]v = v _{i} \: e ^{ - ct}[/tex]
[tex]10= v _{i} \: e ^{ - ct} = (v _{1})(1)[/tex]
[tex] v _{i} = 10 \:m /s[/tex]
At t = 20. seconds.
v = 5 m/s
[tex]5= 10 \: e ^{ - c(20 \: s)}[/tex]
[tex] e ^{ - c(20 \: s)} = 0.500[/tex]
At t = 40 seconds.
[tex]v = 10 \: e ^{ - 40 \: c}[/tex]
[tex] = \frac{10}{0.250} [/tex]
= 2.50 m/s
Acceleration of the motor boat is,
[tex]a = \frac{dv}{dt} [/tex]
[tex] = \frac{d}{dt} \: v = v _{i} \: e ^{ - ct}[/tex]
[tex] = v _{i} \: e ^{ - ct}(-c)[/tex]
[tex]= -c(v _{i} \: e ^{ - ct})[/tex]
a = -cv
The acceleration of the motorboat is negative constant into speed.
Therefore, the acceleration of the motorboat is proportional to the speed.
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(ii) Write down, using figures and unit abbreviations: six thousand watts; four
hundred pascals.
a) 6000 Pa
b) 400 Pa
The watt (abbreviated W) is the International System of Units' (SI) standard unit of power (energy per unit time), the equivalent of one joule per second. The watt is used to specify the rate at which electrical energy is dissipated, or the rate at which electromagnetic energy is radiated, absorbed, or dissipated.
One pascal can be defined as the pressure exerted on a surface of area 1m2 by a force of 1N that acts normally on it.
in statement six thousand watt
six thousand in figure will be = 6000
since , its unit is in watt it can be written as = 6000 W
in statement four hundred pascals
four hundred in figures will be = 400
having unit pascal = Pa
which can be written as = 400 Pa
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the hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. what is the energy of this radiation? (h
The energy of this radiation is 4.58 x 10⁻¹⁹ joules.
What is radiation?
Ionizing and non-ionizing categories of radiation are typically distinguished based on the energy of the radiated particles. Ionizing radiation carries energies greater than 10 eV, which is enough to ionize atoms and molecules and break chemical bonds. This distinction is important since these compounds differ significantly in how poisonous they are to living creatures. Ionizing radiation is typically produced by radioactive materials that release radiation in the form of photons, electrons, or positrons. Other sources include secondary cosmic rays that interact with Earth's atmosphere to produce muons, mesons, positrons, neutrons, and other particles, as well as X-rays from radiography examinations used in medicine.
Calculations:
wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /λ
value for c is 3 x 10⁸ m/s
E is equal to (6.626 x 10 34 J)(3 x 108 m/s) / 434 x 109m.
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules
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The charge per unit length on a long, straight filament is -90.0μC/m . Find the electric field (a) 10.0cm
The electric field for 10 cm is 16.2 MN/C
Length of the straight filament = (10cm) (1m/100cm) = 0.1
Electric field for the straight conductor = [tex]E = \frac{2k_{e} \lambda}{r}[/tex]
E = [2(8.99 × 10^9 N·m^2/C^2)(90.0 × 10^-6 C/m)] / 0.1 m
E = 16.2 MN/C
The electric field is directly radially inward, toward the filament.
Electric field is a force produced by a charge near its surroundings. This force is exerted on other charges when brought in the vicinity of this field. SI unit of electric field is N/C (Force/Charge).
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A car moves 92 km due east, then 56 km due west. what is its total displacement?
By using vector, the total displacement is 36 km.
We need to know about vectors to solve this problem. Vector has magnitude and also direction. Displacement is included in vector. It can be written as
s = sx i + sy j + sz k
where s is displacement, sx is displacement on x-axis, sy is displacement on y-axis and sz us displacement on z-axis.
From the question above, we know that
east = positive x
west = negative x
s1 = 92i km
s2 = -56i km
Hence, the total displacement is
s = s1 + s2
s = 92i - 56i
s = 36i
Find the magnitude
|s| = √sx²
|s| = √36²
|s| = 36 km
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the electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 n/c and directed outward from the axis of the rod. (a) how much charge per unit length exists on the copper rod? (b) what would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
a)The charge per unit length (in C/m) exists on the copper rod is
3.34×10⁻¹² C/m
b) The electric flux through a cube is 0.018 N.m²/C
Electric field = it is a physical field that surrounded by electric charge particle and exert force on another charge particles.
Radius of copper rod = 1 cm. = 0.01m
The electric field strength = 3 N/C.
The electric field at 2 cm from the center of a long copper rod.
a)The charge per unit length (in C/m) exists on the copper rod.
The electric field at a point can be given as
E = λ/2πε₀r
where
λ is linear charge density
ε₀ is the permittivity in free space
r is the radius
λ = E*2πε₀r
= 3 ×2×3.14×8.85×10⁻¹² ×2×10⁻²
λ = 3.34×10⁻¹² C/m
The charge per unit length (in C/m) exists on the copper rod is
λ = 3.34×10⁻¹² C/m
b) side of the cube is 5cm
charge enclosed inside rod is given by :
q = λ L
using the equation of gauss law
Ф = q/ε₀
using this in the above equation we get
q(enclosed) = λ L/ε₀
q = 3.34×10⁻¹² C/m × 5× 10⁻²m / 8.85×10⁻¹²C²/N.m²
= 0.018 N.m²/C
the electric flux through a cube is 0.018 N.m²/C
a)The charge per unit length (in C/m) exists on the copper rod is
3.34×10⁻¹² C/m
b) The electric flux through a cube is 0.018 N.m²/C
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a small turbo prop commuter airplane accelerates before taking off calculate the acceleration of the planbe
The calculated acceleration is for plane and in 19 s 2.63 m/s2 and 475.25 m respectively.
An object's velocity can alter depending on whether it moves faster or slower or in a different direction. A few instances of acceleration include an apple falling, the moon orbiting the earth, or a car being halted at a set of traffic lights.
We employ the following equation to resolve the first portion of this problem.
Vf = 50 m/s final speed
Initial speed: 0 V
t=time
a=acceleration
a=(Vf-Vo)/t
a= (50-0)/19=2.63m/s^2
The following equation is used for this problem's second component.
X=(Vf^2-Vo^2)/2a
X= (50^2-0^2)/ (2*2.63)
=475.25m
The complete question is- A small turbo prop commuter airplane, starting from rest on a New York airport runway, accelerated for 19.0s before taking off. It's speed at takeoff is 50.0 m/s. Calculate the acceleration of the plane, assuming it remains constant. Part B) In this problem, how far did the plane move while accelerating for 19.0s?
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a horizontal clothesline is tied between 2 poles, 10 meters apart. when a mass of 1 kilograms is tied to the middle of the clothesline, it sags a distance of 1 meters. what is the magnitude of the tension on the ends of the clothesline?
By analyzing the vector, the clothesline tension is 25 N.
We need to know about vectors to solve this problem. Force is included in vectors that have magnitude and direction. It can be written as
F = (Fx i + Fy j) N
where F is the force vector, Fx is the x-axis component and Fy is the y-axis component.
From the question we know that :
x = 10 m
(midpoint = 5m)
y = 1 m
m = 1 kg
Find the weight
W = m . g
W = 1 x 10
W = 10N
There are two y-axis components of the tension that hold the clothes. Because of static conditions, we can write
∑Fy = 0
Ty + Ty - W = 0
2Ty - 10 = 0
2Tsinθ - 10 = 0
Find sinθ
sinθ = y/x
sinθ = 1/5
Substitute the sinθ
2Tsinθ - 10 = 0
2T(1/5) - 10 = 0
2T/5 = 10
T = 25 N
Hence, the clothesline tension is 25 N.
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when a 0.30 kg mass is suspended from a massless spring, the spring stretches a distance of 2.0 cm. let 2.0 cm be the rest position for the mass-spring system. the mass is then pulled down an additional distance of 1.5 cm and released.calculate the period of resulting oscillation in si units.
When a 0.30 kg mass is suspended from a massless spring, the spring stretches a distance of 2.0 cm, with an additional distance of 1.5cm,
Mass m= 0.30 kg
Distance x = 2.0 + 1.5 cm = 0.035 m
Since spring constant k = [tex]{\frac{mg}{x}[/tex] , (g=9.81)
k = [tex]{\frac{0.3*9.81}{0.035}[/tex] = 84.085 N/m
ω = [tex]\sqrt{\frac{k}{m} }[/tex] = [tex]\sqrt{\frac{84.085}{0.3} }[/tex] = 16.74 rad/sec
T = [tex]\frac{2\pi }{w}[/tex] = [tex]\frac{2*3.14}{16.74}[/tex] = 0.375 sec
The spring constant is the force required to stretch or compress the spring divided by the distance the spring is lengthened or shortened. It is used to determine the stability or instability of a spring and thus the system for which it is intended.
Therefore the time period of resulting oscillation is 0.375 sec.
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Q|C A bat, moving at 5.00 m/s , is chasing a flying insect. If the bat emits a 40.0 -kHz chirp and receives back an echo at 40.4 kHz , (a) what is the speed of the insect?
The speed of the insect is 3.03m/s.
What is speed?The speed of an object, also known as v in kinematics, is a scalar quantity that refers to the size of the change in that object's position over time or the size of the change in that object's position per unit of time. The instantaneous speed is the upper limit of the average speed as the duration of the time interval gets closer to zero. The average speed of an object over a period of time is calculated by dividing its distance traveled by the length of the interval. Distance divided by time is the dimension of speed.
Explanation:
The frequency that is received by the source after the reflection is given by,
[tex]f_{1} =f_{0} (\frac{V-v}{V-V_{b} } )(\frac{V+v_{b} }{V+v} )[/tex]
[tex]40.4=40 (\frac{343-v}{343-5} ) x (\frac{343+5}{343+v} )\\[/tex]
[tex]v=3.03 m/s[/tex]
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question 53 a 250 ft long span separates a dc power supply from a lamp which draws 25 a of current. if 14 awg wire is used (note that two wires are needed for a total of 500 ft), calculate the amount of power wasted in the wire.
A bulb that consumes 25 A of current is separated from a dc power source by a span of 250 feet. Calculate the amount of power lost in the cable if 14 AWG wire is used (keep in mind that two wires are required for a total length of 500 ft).
How much energy is lost?The answer is 34%, according to the Energy Information Administration (EIA). In other words, by the time the power reaches the consumer metre, 66% of the raw energy needed to generate it has been lost.
The biggest televisions in particular might be the most energy-hungry of all entertainment devices, according to Brian Horne, Senior Insights & Analytics Consultant, Energy Saving Trust. Regardless of its energy rating, a television will use more energy the bigger it is.
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I can’t think of the answer for this question
The car started 15 ml west of Mulberry road.
Consider the East-West route as the x-axis, and place Mulberry Road's intersection at x = 0. This implies that, the west of the intersection is -ve and the east side will be represented positive.
Given the car ends up at 18 mi West to the intersection, implies that it ends at -18mi from x = 0. Also given the total displacement = - 33mi.
Hence, initial position = final position - displacement
= -18 - (- 33)
= -18 + 33
= 15 mi. So, the car started from 15mi from the Mulberry road and from the East of it.
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A person walks 15 km in 2 hour 30 minutes. Calculate his average speed.
Answer:
6 kilometers per hour
Explanation:
Distance = 15km
Time= 2 hours 30 minutes
Speed = Distance / Time
Speed = 15 / 2.5
Speed = 6 km per hour.