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Tatiana [17]
2 years ago
5

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ng the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 4.1 m/s, and 65-kg person feels a 455-N force pressing against his back. What is the radius of a chamber?
Physics
1 answer:
ozzi2 years ago
4 0

Answer:

r=2.4m

Explanation:

We have to use the centripetal force  equation

Fc=\frac{mv^{2} }{r}

we  need the radious so we have to isolate "r" and we get

r=\frac{mv^{2} }{Fc}

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

r=\frac{65*4.1^{2} }{455}

r=2.4m

The radius of the amusement park chamber is 2.4m

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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0
2 years ago
Explain the relationship between the current output of the power supply and the current through each component in the parallel c
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A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
2 years ago
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