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JulsSmile
16 days ago
14

A stagehand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N. W

hat is the coefficient of static friction between the stage floor and the 490 N piece of scenery? a. 0.36 c. 2.8 b. -2.8 d. -0.36
Physics
1 answer:
Softa [2K]16 days ago
6 0
The answer is option (a). A stagehand applies a horizontal force of 176 N to start moving a large piece of scenery that was initially at rest. The equation to determine this force has to do with its rotational speed and centripetal acceleration.
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18 days ago
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
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If my calculations are accurate, the angle is 67.5 degrees.

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1 month ago
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Alicia intends to swim to a point straight across a 100 m wide river with a current that flows at 1.2 m/s. She can swim 2.5 m/s
Sav [2230]

Answer:

θ = 61.3°

Alicia must swim at an angle of 61.3°

Explanation:

Parameters given include:

Width of the river = 100 m

Alicia's speed in still water = 2.5 m/s

Speed of river's current = 1.2 m/s

The angle she needs to swim can be determined by combining the velocities, taking into account the current's influence.

Her swimming speed aimed against the current must offset the current's velocity;

2.5cosθ - 1.2 = 0

2.5cosθ = 1.2

cosθ = 1.2/2.5

θ = cosinverse(1.2/2.5)

θ = 61.3°

4 0
18 days ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
kicyunya [2264]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For the maximum emf, \sin\omega t=1

Therefore,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

Hence, the maximum emf generated in the loop is 5.32 V.

3 0
28 days ago
A 6V radio with a current of 2A is turned on for 5 minutes. Calculate the energy transferred in joules
ValentinkaMS [2425]

Answer:

R=V/I=6/2=3 ohm

time = 5 minutes = 5*60=300 seconds

I=2 A

Energy = I²Rt=(2)²*3*300=4*900=3600 J

7 0
1 month ago
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