Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the first complete revolution, how long will it take to make the next complete revolution?
Answer:
The time taken is [tex]\Delta t = 1.5 \ s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 1.25 \ kg[/tex]
The time taken to make the first complete revolution is t= 3.60 s
The displacement of the first complete revolution is [tex]\theta = 1 rev = 2 \pi \ radian[/tex]
Generally the displacement for one complete revolution is mathematically represented as
[tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]
Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]
[tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]
[tex]\alpha = 0.9698 \ s[/tex]
Now the displacement for two complete revolution is
[tex]\theta_2 = 2 * 2\pi[/tex]
[tex]\theta_2 = 4\pi[/tex]
Generally the displacement for two complete revolution is mathematically represented as
[tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]
=> [tex]t^2 = 25.9187[/tex]
=> [tex]t= 5.1 \ s[/tex]
So
The time taken to complete the next oscillation is mathematically evaluated as
[tex]\Delta t = t_2 - t[/tex]
substituting values
[tex]\Delta t = 5.1 - 3.60[/tex]
[tex]\Delta t = 1.5 \ s[/tex]
The time for the ball to complete the next revolution is 1.5 s.
The given parameters;
mass of the ball, m = 1.25 kgtime of motion, t = 3.6 sone complete revolution, θ = 2πThe constant angular acceleration of the ball is calculated as follows;
[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]
The time to complete the next revolution is calculated as follows;
[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]
[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]
Thus, the time for the ball to complete the next revolution is 1.5 s.
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At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W
Complete Question
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.
Answer
a. Approximately [tex]5*10^{10}[/tex] fissions per second.
b. Approximately [tex]6*10^{12 }[/tex]fissions per second.
c. Approximately [tex]4*10^{11}[/tex] fissions per second.
d. Approximately [tex]3*10^{12}[/tex] fissions per second.
e. Approximately[tex]3*10^{14}[/tex] fissions per second.
Answer:
The correct option is d
Explanation:
From the question we are told that
The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]
Thus to generated [tex]100 \ J/s[/tex] i.e (100 W ) the rate of fission is
[tex]k = \frac{100}{3.2 *10^{-11} }[/tex]
[tex]k =3*10^{12} fission\ per \ second[/tex]
Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
Answer:
T = 3.14 hours
Explanation:
We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
We know that the radius of Mars is 3,389.5 km.
So, r = 1,787 + 3,389.5 = 5176.5 km
Using Kepler's law,
[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]
M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]
So,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]
or
T = 3.14 hours
So, the orbital period is 3.14 hours
Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?
Answer: Only Tech B is correct.
Explanation:
First, tech A is wrong.
The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.
Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.
Answer:
b. The current in the loop always flows in a counterclockwise direction.
Explanation:
When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.
The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.
This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
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B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.
Answer:
(A). 1620 watt.
(B).0.8885.
Explanation:
So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;
Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.
(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).
(B). the rms values to find the power efficiency Pout/Pin of the inverter.
P(in) = 195 × 9.35 = 1823.3 watt.
Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.
A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 960kg . Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift.
Answer:
602.27 kg
Explanation:
The computation of the largest mass of cargo the balloon can lift is shown below:-
Volume of helium inside the ballon= (4 ÷ 3) × π × r^3
= (4 ÷ 3) × 3.14 × 6.953
= 1406.19 m3
Mass the balloon can carry = volume × (density of air-density of helium)
= 1406.19 × (1.29-0.179)
= 1562.27 kg
Mass of cargo it can carry = Mass it can carry - Mass of structure
= 1562.27 - 960
= 602.27 kg
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?
Explanation:
Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,
[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]
[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]
[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]
[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]
after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,
[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]
[tex]\qquad \Phi_{B, f}=0[/tex]
(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,
[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]
[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]
[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]
(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb
(b) The induced EMF in the coil is 0.583 mV
Flux and induced EMF:Given that the coil has N = 250 turns
and an area of A = 14cm² = 1.4×10⁻³m².
It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T
(a) The flux passing through the coil is given by:
Ф = NBAcosθ
where θ is the angle between area vector and the magnetic field
The area vector is perpendicular to the plane of the coil.
So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil
So the initial flux is:
Φ = NABcos0° = NAB
Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb
Ф = 1.75 × 10⁻⁵ Wb
Finally, θ = 90°, and since cos90°, the final flux through the coil is 0
(b) The EMF induced is given by:
E = -ΔФ/Δt
E = -(0 - 1.75 × 10⁻⁵)/0.030
E = 0.583 × 10⁻³ V
E = 0.583 mV
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Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s
Answer:
The time difference is 2.97 sec.
Explanation:
Given that,
Tension = 29 N
Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]
Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]
Length of first section = 2.5 m
Length of second section = 2.8 m
We need to total distance of first pulse
Using formula for distance
[tex]d=2.5+2.5[/tex]
[tex]d_{1}=5.0\ m[/tex]
We need to total distance of second pulse
Using formula for distance
[tex]d=2.8+2.8[/tex]
[tex]d_{2}=5.6\ m[/tex]
We need to calculate the speed of pulse in the first string
Using formula of speed
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]
Put the value into the formula
[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]
[tex]v_{1}=1.24\ m/s[/tex]
We need to calculate the speed of pulse in the second string
Using formula of speed
[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]
[tex]v_{2}=0.80\ m/s[/tex]
We need to calculate the time for first pulse
Using formula of time
[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]
[tex]t_{1}=4.03\ sec[/tex]
We need to calculate the time for second pulse
Using formula of time
[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]
[tex]t_{2}=7\ sec[/tex]
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=t_{2}-t_{1}[/tex]
Put the value into the formula
[tex]\Delta t=7-4.03[/tex]
[tex]\Delta t=2.97\ sec[/tex]
Hence, The time difference is 2.97 sec.
what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor
Answer
The effect is that it Decreases the field current IF and increases slope K1
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N
A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?
Answer:
c. 70 Ω
Explanation:
The R and R resistors are in parallel. The 2R and 2R resistors are in parallel. The 4R and 4R resistors are in parallel. Each parallel combination is in series with each other. Therefore, the equivalent resistance is:
Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)
Req = R/2 + 2R/2 + 4R/2
Req = 3.5R
Req = 70Ω
which category would a person who has an IQ of 84 belong ?
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.
A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?
Answer:
There is no induced current on the coil.
Explanation:
Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.
A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about
Answer:
-50N
Explanation:
F=ma=m(Vf-Vi)/t
m=10kgVf=0m/sVi=10m/st=2sF=(10)(-10)/(2)=-50N
So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is
Complete question is;
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
13 m
Explanation:
We are given;
Distance between two nearly parallel mirrors; d = 6.5 m
Distance between the face and the nearer mirror; x = 3 m
Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m
Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.
Thus;
Distance of the first reflection of the back of the head in the rear mirror from the object head is;
y' = 2y
y' = 2 × 3.5
y' = 7
The total distance of this image from the front mirror would be calculated as;
z = y' + x
z = 7 + 3
z = 10
Finally, the second reflection of this image will be 10 meters inside in the front mirror.
Thus, the total distance of the image of the back of the head in the front mirror from the person will be:
T.D = x + z
T.D = 3 + 10
T.D = 13m
Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet
Answer:
A. magnetic field
Explanation:
The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .
What is a magnetic field ?A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.
As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.
The magnetic field is produced as a result when an electrical current is passed through the conducting wire .
Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .
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What is the reason for the increase and decrease size of the moon and write down in a paragraph.
Answer:
The reason for the increase or decrease of the moon is due to the angular perception of the moon.
Explanation:
Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.
The zenith is the highest part of the sky and the horizon the lowest.
When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.
But when looking up at the sky as there is no reference point there will be a failure in the perception of size.
Give an example of a fad diet that is not healthy and one that is healthy. Explain how you know the difference.
Answer:
Good Diet: ! gallon of water a day, Fruits, Vegetables, White meats(Chicken), Don't eat past 3 PM.
Bad Diet: Pizza, Red meat, Baked goods, Eating at late hours.
Explanation: I know the difference because, When you drink water first thing in the morning it gets your metabolism running. Than means you can digest foods better, you want to feed your body good foods but you should not eat until you feel stuffed. You should eat until you are no longer starving. Than you should drink a cup of water in between meals. I know you should not eat past 3 pm because your body needs time to digest foods because you should never go to sleep with a full stomach. I know the difference between good food and bad food because when you eat healthy food and a balanced diet, your body will have more energy and you wont feel tired afterwards. Eating bad foods and food with artificial sugars will clump up in your kidneys, and your body will have small bursts of energy but you will feel lazy afterwards...Your body is supposed to stay energized from a healthy meal in order to give you the energy your body needs to exercise. If you feel droopy all the time and you don't want to do anything, than you are unhealthy.
Answer:
A vegetarian diet is an example of a good fad diet if you do it correctly. It can help you get lots of veggies and good nutrients from them while still following the non-meat diet you want. This can be effective and good for weight loss becasue you are still eating and getting all the good nutrients and calories from less fatty foods.
Vegan diet (some can be successful but many people fail and do not do good that is why I choose this) The problem with this fad diet is that it can cause nutritional deficiencies and lead to a host of additional health problems, including negatively impacting hormonal health and metabolism. Many people also struggle to find healthy vegan food and end up eating bad and fatty foods instead.
Explanation:
Got a 100
The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
Answer:
Speed of the electron is 2.46 x 10^8 m/s
Explanation:
momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]
where [tex]M_{0}[/tex] is the rest mass of the electron
V is the velocity of the electron.
under relativistic effect, the mass increases.
under relativistic effect, the new mass M will be
M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]
where
[tex]\beta = V/c[/tex]
c is the speed of light = 3 x 10^8 m/s
V is the speed with which the electron travels.
The new momentum will therefore be
==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have
1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
the equation reduces to
1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]
square both sides of the equation, we have
3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]
3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1
2.0625 = 3.0625[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 0.67
β = 0.819
substitute for [tex]\beta = V/c[/tex]
V/c = 0.819
V = c x 0.819
V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s
Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Answer:
Explanation:
magnetic field due to an infinite current carrying conductor
B = k x 2I / r where k = 10⁻⁷ , I is current in conductor and r is distance from wire
putting the given data
B = 10⁻⁷ x 2 x 100 / 8
= 25 x 10⁻⁷ T .
B )
earth's magnetic field = .5 gauss
= .5 x 10⁻⁴ T
= 5 x 10⁻⁵ T
percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100
= 5 %
C )
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?
Answer:
The tension on string when the speed was raised is 134.53 N
Explanation:
Given;
Tension on the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The speed of the wave is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length
[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]
The final tension T₂ will be calculated as;
[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]
Therefore, the tension on string when the speed was raised is 134.53 N
Please help!
Much appreciated!
Answer:
your question answer is 22°
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth
2. Give a specific example of a system with the energy transformation shown.
a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?
Answer:
a
[tex]m = 0.169 \ kg[/tex]
b
[tex]|v_{max} |= 0.5653 \ m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 14 \ N/m[/tex]
The maximum extension of the spring is [tex]A = 6.0 \ cm = 0.06 \ m[/tex]
The number of oscillation is [tex]n = 30[/tex]
The time taken is [tex]t = 20 \ s[/tex]
Generally the the angular speed of this oscillations is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
where T is the period which is mathematically represented as
[tex]T = \frac{t}{n}[/tex]
substituting values
[tex]T = \frac{20}{30 }[/tex]
[tex]T = 0.667 \ s[/tex]
Thus
[tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]
[tex]w = 9.421 \ rad/s[/tex]
this angular speed can also be represented mathematically as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]m =\frac{k }{w^2}[/tex]
substituting values
[tex]m =\frac{ 15 }{(9.421)^2}[/tex]
[tex]m = 0.169 \ kg[/tex]
In SHM (simple harmonic motion )the equation for velocity is mathematically represented as
[tex]v = - Awsin (wt)[/tex]
The velocity is maximum when [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]
[tex]v_{max} = - A* w[/tex]
=> [tex]|v_{max} |= A* w[/tex]
=> [tex]|v_{max} |= 0.06 * 9.421[/tex]
=> [tex]|v_{max} |= 0.5653 \ m/s[/tex]
Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q
Answer:
Upper plate Q/3
Lower plate 2Q/3
Explanation:
See attached file
