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Licemer1 [7]
3 years ago
10

You have a summer job working at a company developing systems to safely lower large loads down ramps. Your team is investigating

a magnetic system by modeling it in the laboratory. The safety system is a conducting bar that slides on two parallel conducting rails that run down the ramp. The bar is perpendicular to the rails and is in contact with them. At the bottom of the ramp, the two rails are connected together. The bar slides down the rails through a vertical uniform magnetic field. The magnetic field is supposed to cause the bar to slide down the ramp at a constant velocity even when friction between the bar and the rails is negligible. Before setting up the laboratory model, your task is to calculate the constant velocity of the bar sliding down the ramp on rails in a vertical magnetic field as a function of the mass of the bar, the strength of the magnetic field, the angle of the ramp from the horizontal, the length of the bar which is the same as the distance between the tracks, and the resistance of the bar. Assume that all of the other conductors in the system have a much smaller resistance than the bar.
If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.

Physics
1 answer:
sweet-ann [11.9K]3 years ago
3 0

Answer:

Explanation:

find the solution below

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A baseball of mass 1.23 kg is thrown at a speed of 65.8 mi/h. What is its kinetic energy?
frosja888 [35]

Given:

The mass of the ball is

m=1.23\text{ kg}

The speed of the ball is

\begin{gathered} v=65.8\text{ mi/h} \\  \end{gathered}

Required: calculate the kinetic energy of the baseball

Explanation: to calculate the kinetic energy of a body we will use the formula as

K.E=\frac{1}{2}mv^2

first, we convert velocity from mi/h into m/s.

we know that

1\text{ mi=1609.34 m}

and

1\text{ h=3600 sec}

then the velocity is

\begin{gathered} v=\frac{65.8\times1602.34\text{ m}}{3600\text{ s}} \\ v=29.29\text{ m/s} \end{gathered}

now plugging all the values in the above formula, we get

\begin{gathered} K.E=\frac{1}{2}mv^2 \\ K.E=\frac{1}{2}\times1.23\text{ kg}\times(29.29\text{ m/s})^2 \\ K.E=527.61\text{ J} \end{gathered}

Thus, the kinetic energy of the baseball is

527.61\text{ J}

4 0
2 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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Answer:

true cuase it is true its not false

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