Answer:
4 x 10⁷m/s
Explanation:
When a charged particle moves in a curved path in a magnetic field, it experiences some magnetic force, and in the absence of any other force, which supplies the centripetal force needed to keep the particle in balance.
Let the magnetic force be [tex]F_{M}[/tex]
Let the centripetal force be [tex]F_{C}[/tex]
=> [tex]F_{M}[/tex] = [tex]F_{C}[/tex] --------------(i)
We know that;
[tex]F_{M}[/tex] = qvBsinθ
Where;
q = charge on the particle
v = speed of the particle
B = magnetic field intensity
θ = angle between the speed and magnetic field vectors
Also;
[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex]
Where;
m = mass of the particle
v = velocity/speed of the particle
r = radius of the circular path of motion.
From equation (i)
qvBsinθ = [tex]\frac{mv^2}{r}[/tex] [divide both sides by v]
qBsinθ = [tex]\frac{mv}{r}[/tex] [make v subject of the formula]
v = qrBsinθ / m --------------------(ii)
From the question;
B = 1.0T
r = 0.4m
θ = 90° [since magnetic field is always perpendicular to velocity]
q = 1.6 x 10⁻¹⁹C [charge of a proton]
m = 1.6 x 10⁻²⁷kg [mass of a proton]
Substitute these values into equation(ii) as follows;
v = (1.6 x 10⁻¹⁹ x 0.4 x 1.0 x sin90°) / (1.6 x 10⁻²⁷)
v = 4 x 10⁷ m/s
Therefore the speed of the proton is 4 x 10⁷m/s
A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system
Answer:
E = 0.645J
Explanation:
In order to calculate the total mechanical energy of the system, you take into account that if the zero of energy is at the equilibrium position, then the total mechanical energy is only the elastic potential energy of the spring.
You use the following formula:
[tex]E=U_e=\frac{1}{2}kA^2[/tex] (1)
k: spring constant = ?
A: amplitude of the oscillation = 7.50cm = 0.075m
The spring constant is given by:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
[tex]k=4\pi^2f^2m[/tex] (2)
f: frequency of the oscillation = 1.95Hz
m: mass of the piece of iron = 1.53kg
You replace the expression (1) into the equation (2) and replace the values of all parameters:
[tex]E=\frac{1}{2}(4\pi^2f^2m)A^2=2\pi^2f^2mA^2\\\\E=2\pi^2(1.95Hz)^2(1.53kg)(0.075m)^2=0.645J[/tex]
The totoal mechanical energy of the system is 0.645J
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other? What would the general shape of the field lines look like? What would the field lines look like in between the two pieces?
Answer:
Explanation:
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a playground merry-go-round of radius r = 2.20 m has a moment of inertia i = 245 kg · m2 and is rotating at 11.0 rev/min about a frictionless vertical axle. facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round?
Answer:
8.92 rpm
Explanation:
Given that
Radius of the merry go round, r = 2.2 m
Initial moment of inertia, I1 = 245 kgm²
Initial speed of rotation, w1 = 11 rpm
Mass of the child, m = 26 kg
To solve the problem, we use the law conservation of momentum
I1w1 = I2w2
I2 = mr² + I1
I2 = 245 + 26 * 2.2
I2 = 245 + 57.2
I2 = 302.2 kgm²
Now, applying the formula, we have
I1w1 = I2w2
245 * 11 = 302.2 * w2
w2 = 2695 / 302.2
w2 = 8.92 rpm
Thus, the new angular speed of the merry go round is 8.92 rpm
A small glass bead charged to 8.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 940 μN.
Required:
What is the total charge on the rod?
Answer:
71nC is the total charge of the rod
Explanation:
See attached file
The total charge on the rod is equal to 3.3 × 10⁻⁸ C.
What is the force on a charge in an electric field?The force on the charge in a uniform electric field E is given by:
F = qE where q is charge in coulombs
The electric field due to the charge associated with the rod is given by:
[tex]E =\frac{kQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]
Where r is the distance between the bead and the rod, L is the length of the glass rod and Q is the charge on the rod.
The force experienced by the bead charged is,
[tex]F =\frac{kqQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]
From the above equation, we can find the value of Q as:
[tex]Q =\frac{Fr\sqrt{r+\frac{L^2}{4} } }{kq}[/tex]
Given, the value of force, F = 940μN = 940 ×10⁻⁹ C
The length of glass rod, L = 10cm = 0.1 m and r = 4cm = 0.04 m
[tex]Q =\frac{(940\times 10^{-6}N)(0.04m)\sqrt{(0.04m)+\frac{(0.1m)^2}{4} } }{(8.99\times 10^9 N.m^2/C^2)(8\times 10^{-9}C)}[/tex]
[tex]Q= 0.5228\times 10^{-6}\times\sqrt{0.0041}[/tex]
[tex]Q = 3.3\times 10^{-8} C[/tex]
Therefore, the total charge on the rod is 3.3 × 10⁻⁸ C.
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Which statement describes a disadvantage of using natural gas as an energy source? It is expensive to use. It is hard to stop using. It is a renewable resource. It is scarce in some parts of the world.
Answer:
B : It is hard to stop using.
Explanation:
just took the quiz ! hope this helps with anyone who needs it !
Due to the dependency on natural gas as a fuel, it is hard to stop using.
What is natural gas?Natural gas is a fossil fuel which is obtained from the ground in association with petroleum.
Natural gas consists mainly of petroleum.
It is a non-renewable energy source.
Natural gas use contributes to global warming
However, due to the dependency on natural gas as a fuel, it is hard to stop using.
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Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?
Answer:
a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Explanation:
a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:
First arrow
[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
Now, the system is expanded and simplified:
[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]
[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]
Where:
[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.
[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
The initial vertical speed of the arrow is:
[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]
Where:
[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.
[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:
[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]
[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{2} - y_{1} = 56.712\,m[/tex]
Second arrow
[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]
[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{3} - y_{1} = 342.816\,m[/tex]
The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.
b) The total energy of each system is determined hereafter:
First arrow
The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:
[tex]E = U + K_{x}[/tex]
The expression is now expanded:
[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]
Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.
[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:
[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]
[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]
If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 201.720\,J[/tex]
Second arrow:
The total mechanical energy is equal to the potential gravitational energy. That is:
[tex]E = m\cdot g \cdot y_{max}[/tex]
[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]
[tex]E = 201.720\,J[/tex]
Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.
By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
What is experiment ?An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.
What is hydrochloric acid?Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.
Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.
By the experiment "By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
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Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping. B. The increase in amplitude of an oscillation by a driving force is called forced oscillation. C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced. D. An oscillation that is maintained by a driving force is called forced oscillation.
Answer:
right A, B, C, D
Explanation:
They ask which statements are true
A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping
B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation
C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.
D) Right. It is a resonant process the driving force keeps the oscillation of the system
Statements that are right as regards oscillation are:
A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.
B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.
C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.
D. An oscillation that is maintained by a driving force is called forced oscillation.
Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.When there is increase in amplitude of an oscillation as a result of driving force then it is termed forced oscillation.Therefore, the options are correct.
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Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 50 centimeters to 80 centimeters
Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
What must be the diameter of a cylindrical 120-m long metal wire if its resistance is to be ? The resistivity of this metal is 1.68 × 10-8 Ω • m.
Answer:
The diameter is [tex]d = 6.5 *10^{-4} \ m[/tex]
Explanation:
From the question we are told that
The length of the cylinder is [tex]l = 120 \ m[/tex]
The resistance is [tex]\ 6.0\ \Omega[/tex]
The resistivity of the metal is [tex]\rho = 1.68 *10^{-8} \ \Omega \cdot m[/tex]
Generally the resistance of the cylindrical wire is mathematically represented as
[tex]R = \rho \frac{l}{A }[/tex]
The cross-sectional area of the cylindrical wire is
[tex]A = \frac{\pi d^2}{4}[/tex]
Where d is the diameter, so
[tex]R = \rho \frac{l}{\frac{\pi d^2}{4 } }[/tex]
=> [tex]d = \sqrt{ \rho* \frac{4 * l }{\pi * R } }[/tex]
[tex]d = \sqrt{ 1.68 *10 ^{-8}* \frac{4 * 120 }{3.142 * 6 } }[/tex]
[tex]d = 6.5 *10^{-4} \ m[/tex]
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50×10^6 m/s. The proton comes momentarily to rest at a distance 5.31×10^(−13)m from the center of the target nucleus, then flies back in the direction from which it came. What is the number of the protons the nucleus has? Assume no electron cloud is there, ε0=8.85x10^(-12) C^2/(Nm^2)
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass [tex]1.67 \times 10^(-27)[/tex]kg, charge +e = [tex]+1.60 \times 10^(-19)[/tex] C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed [tex]2.50 \times 10^6[/tex] m/s. The proton comes momentarily to rest at a distance [tex]5.31 \times 10^(-13)[/tex] m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are [tex]5.31 \times 10^{-13}[/tex] m apart?
Explanation:
The given data is as follows.
Mass of proton = [tex]1.67 \times 10^{-27}[/tex] kg
Charge of proton = [tex]1.6 \times 10^{-19} C[/tex]
Speed of proton = [tex]2.50 \times 10^{6} m/s[/tex]
Distance traveled = [tex]5.31 \times 10^{-13} m[/tex]
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
[tex](K.E + P.E)_{initial}[/tex] = [tex](K.E + P.E)_{final}[/tex]
[tex](\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)[/tex]
where, [tex]\frac{kq_{p}q_{t}}{r} = U = Electric potential energy[/tex]
U = [tex](\frac{1}{2}m_{p}v^{2}_{p})[/tex]
Putting the given values into the above formula as follows.
U = [tex](\frac{1}{2}m_{p}v^{2}_{p})[/tex]
= [tex](\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})[/tex]
= [tex]5.218 \times 10^{-15} J[/tex]
Therefore, we can conclude that the electric potential energy of the proton and nucleus is [tex]5.218 \times 10^{-15} J[/tex].
When we describe electric flux, we say that a surface is oriented in a certain direction with respect to an electric field. When we try to calculate how much electric field passes through the surface, we make use of the:_________.
1. Wedge Product
2. Dot Product
3. Cross Product
Answer:
2. Dot Product
Explanation:
The calculation of the electric flux gives an scalar result.
When we tray to calculate how much electric field passes trough a surface, we are calculating a scalar value. Furthermore, the concept of flux requires the calculation of a scalar value.
Also it is necessary to take into account that the magnitude of the flux trough a surface depends of the inclination of the surface respect to the direction of the electric field. This is taken into account sufficiently by a dot product.
Then, the answer is:
2. Dot Product
You push very hard on a heavy desk, trying to move it. You do work on the desk: You push very hard on a heavy desk, trying to move it. You do work on the desk: whether or not it moves, as long as you are exerting a force. only if it starts moving. never-it does work on you. only if it doesn't move. None of the above.
Answer:
Only if it starts moving
Explanation:
Work done is defines as [tex]W=F.d=Fdcos\Theta[/tex]
In two case work done will be zero
First case is that when force and displacement are perpendicular to each other
What's more, other case is that when there is no displacement
So for work to be done there must have displacement, in the event that there is no displacement then there is no work done
Two moons orbit a planet in nearly circular orbits. Moon A has orbital radius r, and moon B has orbital radius 16r. Moon A takes 10 days to complete one orbit. How long does it take moon B to complete an orbit
Answer:
Kepler's Third Law: The square of the period of any planet about the sun is proportional to cube of its mean distance from the sun.
Mathematically: T^2 = K R^3
So (TA / TB)^2 = (RA / RB)^3
TB^2 = TA^2 * (RB / RA)^3
TB^2 = 10^2 * 16^3
TB = (409600)^1/2 = 640 days
A uniform electric field E = E0 j is set-up in a region of space. A frame is placed in that region in such a way that its plane is perpendicular to the y-axis. Which of the following changes would decrease the magnitude of the electric flux through the frame?A) sliding the frame sideways parallel to the x-axis within the xz-planeB) Sliding the frame sideways parallel to the z-axis within the xz-planeC) moving the frame vertically along the y-axis keeping parallel to the xz-planeD) rotating the frame in the xz-plane with respect to the y-axisE) tilting the frame so that its plane is now in the yz-plane
Answer:
The correct answer to the following question will be Option E (Leaning back the camera such that the plane seems to be in the yz-axis now.).
Explanation:
The given field is:
⇒ [tex]E=EO \ j[/tex]
The above-given field does have an E-field perpendicular to something like the plane.
So,
⇒ [tex]ECos\theta=ECos 90^{\circ}[/tex]
We know that the value of "Cos 90°" is zero.
⇒ [tex]Cos 90^{\circ}=0[/tex]
The other given choices are not related to the given circumstances. So that Option E seems to be the right answer.
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
Answer:
Explanation:
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
A ball bouncing against the ground and rebounding is an example of an elastic collision. Describe two different methods of evaluating this interaction, one for which momentum is conserved, and one for which momentum is not conserved. Explain your answer.
Answer:
Momentum is conserved when there are no outside forced present and it has an equal and opposite reaction, also momentum is conserved the ball's momentum is transferred to the ground. This first instance is the case of a Closed system.
The second case where momentum is not conserved is when there is a variation or difference in the moment of the ball because of influence of external forces
An electron traveling north enters a region where the electric field is uniform and points east. Will the electron speeds up or slows down?
Answer:
electron has a constant velocity in the north direction and accelerates in the west direction, the electron is accelerating and its velocity on this axis increasing in the west direction.
Explanation:
In this exercise we are asked about the speed of the electron
The electric force is F = q E
the charge of the electron is q = -e = - 1.6 10-19 C
If we use Newton's second law
Y axis (North-south direction)
[tex]v_{y}[/tex] = cte
X axis (East-West direction)
F = m a
- e E = m a
a = - e / m E
whereby electron has a constant velocity in the north direction and accelerates in the west direction, the electron is accelerating and its velocity on this axis increasing in the west direction.
A factory worker moves a 24.0 kg crate a distance of 4.90 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.200.
1. What magnitude of force must the worker apply?
2. How much work is done on the crate by the worker's push?
3. How much work is done on the crate by friction?
4. How much work is done by normal force?
5. What is the net work done on the crate?
Answer:
1) The magnitude of force that the worker must apply is 47.074 newtons, 2) The work done by the worker's push is 230.663 joules, 3) The work done by friction is -230.661 joules, 4) The work done by normal force is 0 joules, 5) The net work done on the crate is 0 joules.
Explanation:
1) The magnitude of the force that worker must apply must be equal to the kinetic friction force between the crate and the floor. (Newton's Third Law) As the box moves at constant velocity, net force of the system must be zero. (Newton's Second Law). The equations of equilibrium for the crate are, respectively:
(Parallel to ground)
[tex]\Sigma F_{x} = P - \mu_{k}\cdot N = 0[/tex]
(Perpendicular to ground)
[tex]\Sigma F_{y} = N - m\cdot g = 0[/tex]
Where:
[tex]P[/tex] - Force exerted on the crate by the worker, measured in newtons.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]\mu_{k}[/tex] - Coefficient of kinetic friction, dimensionless.
The magnitude of the force is obtained after reducing the system of equations and replacing every known variable:
[tex]P = \mu_{k}\cdot m \cdot g[/tex]
If [tex]\mu_{k} = 0.200[/tex], [tex]m = 24\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the magnitude of the force is:
[tex]P = (0.200)\cdot \left(24\,kg\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]P = 47.074\,N[/tex]
The magnitude of force that the worker must apply is 47.074 newtons.
2) As force is constant and parallel to motion, the work done on the crate by the worker is calculated by multiplying the force by the travelled distance:
[tex]W_{P} = (47.074\,m)\cdot (4.90\,m)[/tex]
[tex]W_{P} = 230.663\,J[/tex]
The work done by the worker's push is 230.663 joules.
3) Friction force is constant and antiparallel to motion. The work done by the friction is equal to the product of the friction force and the travelled distance. That is:
[tex]W_{f} = -(0.200)\cdot (24\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.90\,m)[/tex]
[tex]W_{f} = -230.661\,J[/tex]
The work done by friction is -230.661 joules.
4) Since normal force is perpendicular to motion. The work done by the normal force is zero.
[tex]W_{N} = 0\,J[/tex]
The work done by normal force is 0 joules.
5) The net work done on the crate is equal to the following sum:
[tex]W_{net} = W_{P} + W_{f} + W_{N}[/tex]
If [tex]W_{P} = 230.663\,J[/tex], [tex]W_{f} = -230.661\,J[/tex] and [tex]W_{N} = 0\,J[/tex], the net work done on the crate is:
[tex]W_{net} = 230.663\,J - 230.661\,J + 0\,J[/tex]
[tex]W_{net} = 0\,J[/tex]
The net work done on the crate is 0 joules.
A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is
Answer:
The kinetic energy at a displacement of half the amplitude is 37.5 J
Explanation:
Given;
total energy on the spring, E = 50 J
When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.
E = K + U
Where;
K is the kinetic energy
U is the elastic potential energy
K = E - U
K = E - ¹/₂KA²
When the displacement is half = ¹/₂(A) = A/₂
K = E - ¹/₂K(A/₂)²
K = E - ¹/₂K(A²/₄)
K = E - ¹₄(¹/₂KA²)
Recall, E = ¹/₂KA²
K = ¹/₂KA² - ¹₄(¹/₂KA²) (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)
K = 1(¹/₂KA²) - ¹₄(¹/₂KA²) = ³/₄(¹/₂KA²)
K = ³/₄(¹/₂KA²)
But E = ¹/₂KA² = 50J
K = ³/₄ (50J)
K = 37.5 J
Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J
The kinetic energy when the displacement is half the amplitude
Given the following data:
Total energy = 50 Joules.Displacement, x = [tex]\frac{A}{2}[/tex]To find the kinetic energy when the displacement is half the amplitude:
The total energy of the system of a block and a spring is the sum of the spring's elastic potential energy and kinetic energy of the block and it's proportional to the square of the amplitude.
Mathematically, the total energy of the system of a block and a spring is given by the formula:
[tex]T.E = U + K.E[/tex] .....equation 1.
[tex]T.E = \frac{1}{2} kA^2[/tex]
Where:
T.E is the total energy.U is the elastic potential energy.K.E is the kinetic energy.A is the amplitude.Making K.E the subject of formula, we have:
[tex]K.E = T.E - U[/tex] .....equation 2.
But, [tex]U = \frac{1}{2} kx^2[/tex] ....equation 3.
Where:
k is spring constant.x is change in position (displacement).Substituting the eqn 3 into eqn 2, we have:
[tex]K.E = T.E - \frac{1}{2} kx^2[/tex]
[tex]K.E = T.E - \frac{1}{2} k(\frac{A}{2})^2\\\\K.E = T.E - \frac{1}{2} k(\frac{A^2}{4})\\\\K.E = T.E - \frac{1}{4} (\frac{1}{2} kA^2)\\\\K.E = T.E - \frac{1}{4} (T.E)\\\\K.E = 50 - \frac{1}{4} (50)\\\\K.E = 50 - 12.5[/tex]
K.E = 37.5 Joules.
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18. Two spaceships are traveling in the same direction. With respect to an inertial frame of reference, spaceship A has a speed of 0.900c. With respect to the same inertial frame, spaceship B has a speed of 0.500c. Find the speed vAB of spaceship A relative to spaceship B.
Answer:
[tex]v_{A} -v_{B}=v_{AB}\\v_{AB} =0.9c-0.5c\\v_{AB}=0.4c[/tex]
Explanation:
Because the spaceships are traveling in the same direction, you should subtract the speed of spaceship A from the speed of spaceship B in order to calculate [tex]v_{AB}[/tex].
A small Keplerian telescope has an objective with a 1.33 m focal length. Its eyepiece is a 2.82 cm focal length lens. It is used to look at a 25000 km diameter sunspot on the sun, a distance 1.5*108 km from Earth.
Required:
What angle is subtended by the sunspot's telescopic image in degree?
Answer:
The angle is [tex]\phi = 0.45 0 ^o[/tex]
Explanation:
From the question we are told that
The objective focal length [tex]f = 1.33 \ m[/tex]
The eyepiece focal length is [tex]f_o = 2.82 \ cm = 0.0282 \ m[/tex]
The diameter of the sunlight is [tex]d = 25000km = 2.5 *10^{7} \ m[/tex]
The distance of the sun from from the earth is [tex]D = 1.5 *10^8 km = 1.5 *10^{11} \ m[/tex]
Generally the magnification of the object is mathematically evaluated as
[tex]m = -\frac{f_o }{f_e }[/tex]
The negative sign is because the lens of the telescope is diverging light
substituting values
[tex]m = -\frac{1.33 }{0.0282 }[/tex]
[tex]m = - 47.16 3[/tex]
Now we can obtain the angle made by the object (sunlight ) with respect to the telescope as follows
[tex]tan \theta = \frac{d}{D}[/tex]
substituting values
[tex]tan \theta = \frac{2.5 *10^{7}}{1.5*10^{11}}[/tex]
[tex]tan \theta = 0.0001667[/tex]
[tex]\theta= tan^{-1}[0.0001667][/tex]
[tex]\theta= 0.00955^o[/tex]
The magnification can also be mathematically represented as
[tex]m = \frac{\phi }{\theta }[/tex]
Where [tex]\phi[/tex] is the angle the image made with telescope
Since the negative sign indicate direction of light movement we will remove it from the calculation below
=> [tex]47.163 = \frac{\phi}{0.00955}[/tex]
=> [tex]\phi = 0.45 0 ^o[/tex]
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?
Answer:
14.1 m/s
Explanation:
From the question,
μk = a/g...................... Equation 1
Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.
make a the subject of the equation
a = μk(g).................. Equation 2
Given: μk = 0.160, g = 9.8 m/s²
Substitute into equation 2
a = 0.16(9.8)
a = 1.568 m/s²
Using,
F = ma
Where F = force, m = mass.
Make m the subject of the equation
m = F/a................... Equation 3
m = 160/1.568
m = 102.04 kg.
Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.
Assuming the initial velocity of the skier to be zero
Fd+mgμ = 1/2mv²........................Equation 4
Where v = speed of the skier after crossing the patch, d = distance/width of the patch.
v = √2(Fd+mgμ)/m)................ Equation 5
Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16
Substitute these values into equation 5
v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]
v = √197.57
v = 14.1 m/s
v = 9.86 m/s
If an ocean wave takes 2 minutes to generate one full cycle, from crest to crest, then we will call this the frequency of the wave. What would its period be
Answer:
Explanation:
The period of the wave will be 2 minutes
frequency = 1 / period
= 1 / 2
= .5 per minute
A ball is thrown horizontally from the top of a 41 m vertical cliff and lands 112 m from the base of the cliff. How fast is the ball thrown horizontally from the top of the cliff?
Answer:
4.78 second
Explanation:
given data
vertical cliff = 41 m
height = 112 m
solution
we know here time taken to fall vertically from the cliff = time taken to move horizontally ..........................1
so we use here vertical component of ball
and that is accelerated motion with initial velocity = 0
so we can solve for it as
height = 0.5 × g × t² ........................2
put here value
112 = 0.5 × 9.8 × t²
solve it we get
t² = 22.857
t = 4.78 second
ball thrown horizontally from the top of the cliff in 4.78 second
A 0.140-kg baseball is thrown with a velocity of 27.1 m/sIt is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?
Answer:
The bat and the ball were in contact for 1.8 x 10⁻³ s
Explanation:
Given;
mass of baseball, m = 0.14 kg
initial velocity of the baseball, u = 27.1 m/s
applied force in opposite direction, F = -5000 N
final velocity in opposite direction, v = -37 m/s
Note: The applied force and final velocity are negative because they act in opposite direction to the initial velocity.
impulse received by the body = change in momentum of the body
Ft = Δmv
Ft = mv - mu
Ft = m(v-u)
t = m(v-u) / F
[tex]t = \frac{0.14(-37-27.1)}{-5000} \\\\t = \frac{0.14(-64.1)}{-5000} \\\\t = \frac{-8.974}{-5000} \\\\t = 0.0018 \ s\\\\t = 1.8*10^{-3} \ s[/tex]
Therefore, the bat and the ball were in contact for 1.8 x 10⁻³ s
g A point mass of 1.5kg is attached to a spring and set to oscillate through simple harmonic oscillations. If the period of the oscillation is 10s, find the spring constant.
Answer:
k = 0.6 N/m
Explanation:
The time period of a spring mass oscillation system is given by the following formula:
T = 2π√(m/k)
where,
T = Time Period of Oscillation = 10 s
m = Mass attached to the spring = 1.5 kg
k = spring constant = ?
Therefore,
10 s = 2π√(1.5 kg/k)
squaring on both sides we get:
100 s² = 4π²(1.5 kg/k)
k = 6π² kg/100 s²
k = 0.6 N/m
A 24-cm circumference loop of wire has a resistance of 0.14 Ω. The loop is placed between the poles of an electromagnet, and a field of 0.55 T is switched on in a time of 15 ms. What is the induced current in the loop?
Answer:
The induced current in the loop is 1.2 A
Explanation:
Given;
length of the wire, L = 24 cm = 0.24 m
resistance of the wire, R = 0.14 Ω.
magnetic field strength, B = 0.55 T
time, t = 15 ms = 15 x 10⁻³ s
Circumference of a circle is given as;
L = 2πr
0.24 = 2πr
r = 0.24 / 2π
r = 0.0382 m
Area of a loop is given as;
A = πr²
A = π (0.0382)²
A = 0.004585 m²
Induced emf is given as;
[tex]emf = -\frac{\delta \phi}{dt}[/tex]
Ф = ΔB x A
Ф = ( 0 - 0.55 T) x 0.004585 m²
Ф = -0.002522 T.m²
[tex]emf = - (\frac{-0.002522}{15*10^{-3}} )\\\\emf = 0.168 \ V[/tex]
According to ohm's law;
V = IR
Where;
I is current
The induced current in the loop is calculated as;
I = V / R
I = 0.168 / 0.14
I = 1.2 A
Therefore, the induced current in the loop is 1.2 A
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
The speed of the arrow after passing through the target is 30.1 meters per second.
Explanation:
The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:
[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]
Where:
[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.
[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.
[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.
The final speed of the arrow is now cleared:
[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]
[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]
If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:
[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]
[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]
The speed of the arrow after passing through the target is 30.1 meters per second.