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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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Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder
kramer

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here  when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

U = \frac{1}{2}kx^2

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

KE = \frac{1}{2}kx^2

here we have

k = 70 N/m

x = 0.4 m

KE = \frac{1}{2}(70)(0.4)^2

KE = 5.6 J

Part c)

Now to find the speed we know that

KE = \frac{1}{2} mv^2

5.6 = \frac{1}{2}0.3v^2

v = 6.11 m/s

so its speed is 6.11 m/s

8 0
3 years ago
Read 2 more answers
How much positive and negative charges is there in a cup of water?
Rina8888 [55]
In a cup of water it is a positive and negative charge of zero
3 0
3 years ago
F an object has a mass of 200 kg and a weight of 1000 N, what is g?
jeka94

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

g = 5 m/s square

7 0
3 years ago
11. A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed i
marshall27 [118]

Answer:

A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer

5 0
3 years ago
A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as
Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

T= 3600\ sec

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\dfrac{2\pi}{T}

Put the value into the formula

\omega=\dfrac{2\pi}{3600}

\omega=0.001745\ rad/s

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

a=r\omega^2

Put the value into the formula

a=0.55\times(0.001745)^2

a=1.675\times10^{-6}\ m/s^2

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

3 0
3 years ago
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