Answer:
Explanation:
Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction , so their relative velocity
= u + v = 2.3 m /s ( given )
We shall apply conservation of momentum law for the movement of astronaut and metal plate
mu = M v where m is mass of astronaut , M is mass of metal plate
71 u = 230 x v
71 ( 2.3 - v ) = 230 v
163.3 = 301 v
v = .54 m / s
u = 1.76 m / s
honeycomb will be at rest because honeycomb surface is frictionless . Plate will slip over it . Over plate astronaut is walking .
a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s
b ) velocity of astronaut relative to honeycomb will be + .54 m /s
Here + ve direction is assumed to be the direction of astronaut .
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.
Hahahahaha. Okay.
So basically , force is equal to mass into acceleration.
F=ma
so when F=ma , we get acceleration=6m/s/s
Force is doubled.
Mass is 1/3 times original.
2F=1/3ma
Now , we rearrange , and we get 6F=ma
So , now for 6 times the original force , we get 6 times the initial acceleration.
So new acceleration = 6*6= 36m/s/s
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)
Answer:
Power=50.17dioptre
Power=50.17D
Explanation:
P=1/f = 1/d₀ + 1/d₁
Where d₀ = the eye's lens and the object distance= 5.70m=
d₁= the eye's lens and the image distance= 0.02m
f= focal length of the lense of the eye
We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m
Therefore, we can calculate the power using above formula
P= 1/5.70 + 1/0.02
Power=50.17dioptre
Therefore, the power the eye's is using to see the object from distance is 5.70D
A skater spins at 3rev/s when she stretches her arms outward. If she keeps her fists on her chest she can spin at 4.5rev/s and her body inertia is 3kg.m2. What is her body inertia when she stretches her arms outward?
Answer:
Body inertia I = 4.5 kg/m^2
Explanation:
Here, we want to calculate the body inertia when the arms are stretched outwards.
We know from the question that angular momentum is conserved
Thus;
I * 3 = 4.5 * 3
I = 4.5 kg/m^2
Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a stopwatch to find the time it takes the ball to roll each distance
Answer:
The Independent variable in this experiment is the time taken by the ball to roll down each distance.
The dependent variable is the distance through which the ball rolls
The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.
Explanation:
The complete question is
Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?
Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.
A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.
The relationship between the dependent and independent variables in an experiment is given as
y = f(x)
where y is the output or the dependent variable,
and x is the independent variable.
Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.
If the magnetic field steadily decreases from BBB to zero during a time interval ttt, what is the magnitude III of the induced current
Answer:
Using ohms law
The current is found from Ohm's Law.
I = V /R = E /R = Bxy /Rt.
A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s
Answer:
C) 200 m/s
Explanation:
The sound travels a total distance of 360 m in 0.03 minutes.
v = (360 m) / (0.03 min × 60 s/min)
v = 200 m/s
A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.
Answer:
Maximum acceleration is 800m/s^2
Explanation:
See attached file
Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy U from the light. The linear momentum these objects will receive is such that:
Answer:
absorbent p = S / c
reflective p = 2S/c
Explanation:
The moment of radiation on a surface is
p = U / c
where U is the energy and c is the speed of light.
In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.
p = S / c
in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process
p = 2S/c
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz
If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Answer:
Infrared telescope and camera
Explanation:
An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.
Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength is 4.0 mV/m at a point 1.5 m away from the center of the circle. At what rate is the magnetic field changing?
Answer:
The rate at which the magnetic field changes is [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 4.0 mV/m = 4.0 *10^{-3} V/m[/tex]
The radius of the circular region where the electric field is induced is
[tex]d = 1.5 \ m[/tex]
Generally the induced electric field is mathematically represented as
[tex]E = - \frac{r}{2} * \frac{\Delta B }{\Delta t }[/tex]
The negative sign show that the induced electric field is acting in opposite direction to the change in magnetic field
Where [tex]\frac{\Delta B }{\Delta t }[/tex] is the change in magnetic field
So
[tex]\frac{\Delta B }{\Delta t } = - \frac{2 * E }{r}[/tex]
substituting values
[tex]\frac{\Delta B }{\Delta t } = - \frac{2 * 4.0 *10^{-3}}{ 1.5 }[/tex]
[tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]
Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g
Answer:
Explanation:
The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .
When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .
The greatest acceleration is attained at maximum displacement or at one of the two extreme end .
Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .
Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .
Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.
Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)
Answer:
The induced emf is [tex]\epsilon =7.68 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]
The initial magnetic field is [tex]B_i = 0.14 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2 }[/tex]
substituting values
[tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]
[tex]r = 1.1*10^{-2} \ m[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]
Where [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as
[tex]d\phi = dB * A * cos\theta[/tex]
=> [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]
Here [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field
Also A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]
[tex]A = 3.8 *10^{-4] \ m^2[/tex]
So
[tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]
[tex]d\phi = -5.32*10^{-5} \ weber[/tex]
So
[tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]
[tex]\epsilon =7.68 \ V[/tex]
A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
g a conductor consists of an infinite number of adjacent wires, each infinitely long. If there are n wires per unit length, what is the magnitude of B~
Answer:
B=uonI/2
Explanation:
See attached file
Exercise 1 - Questions 1. Hold the grating several inches from your face, at an angle. Look at the grating that you will be using. Record what details you see at the grating surface. 0 Words 2. Hold the diffraction grating up to your eye and look through it. Record what you see. Be specific. 0 Words 3. Before mounting the diffraction grating, look through the opening that you made for your grating. Record what you see across the back of your spectroscope.
Answer:
1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
2)If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum
3) must be able to see the well-collimated light emission source
Explanation:
1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.
These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.
The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.
2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.
If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.
3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.
The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home
Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
"A power of 200 kW is delivered by power lines with 48,000 V difference between them. Calculate the current, in amps, in these lines."
Answer:
9.6×10⁹ A
Explanation:
From the question above,
P = VI.................... Equation 1
Where P = Electric power, V = Voltage, I = current.
make I the subject of the equation
I = P/V............. Equation 2
Given: P = 200 kW = 200×10³ W, V = 48000 V.
Substitute these vales into equation 2
I = 200×10³×48000
I = 9.6×10⁹ A.
Hence the current in the line is 9.6×10⁹ A.
A proton moving at 4.80 106 m/s through a magnetic field of magnitude 1.74 T experiences a magnetic force of magnitude 7.00 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)
Answer:
31.55° and 148.45°
Explanation:
Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;
F = qvBsinθ where;
F is the magnetic force experienced by the proton
q is the charge on the proton
v is the velocity of the proton
B is the magnetic field
θ is the angle between the proton's velocity and the field (Required)
Given parameters
F = 7.00 * 10⁻¹³N
q = 1.602*10⁻¹⁹C
v = 4.80 * 10⁶ m/s
B = 1.74 T
θ =?
From the formula F = qvBsinθ;
sinθ = F/qvB
sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74
sinθ = 7.00 * 10⁻¹³/13.38*10⁻¹³
sinθ = 0.5231689 * 10⁰
sinθ = 0.5231689
θ = sin⁻¹0.5231689
θ = 31.55°
The following are the positive values of the angle between 0° and 360°
Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.
Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°