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Lisa [10]
3 years ago
8

A small disk of radius R1 is mounted coaxially with a larger disk of radius R2. The disks are securely fastened to each other an

d the combination is free to rotate on a fixed axle that is perpendicular to a horizontal frictionless table top, as shown in the overhead view below. The rotational inertia of the combination is I. A string is wrapped around the larger disk and attached to a block of mass m, on the table. Another string is wrapped around the smaller disk and is pulled with a force F as shown. The acceleration of the block is:
Physics
1 answer:
murzikaleks [220]3 years ago
3 0

Answer:

Explanation:

For moment of inertia, the mass will appear as a point mass on the rim of the larger radius.

Total moment of inertia of the system is (I + mR₂²)

torque is FR₁

α = τ/I = FR₁/(I + mR₂²)

a = R₂α = R₂(FR₁/(I + mR₂²))

a = FR₁R₂/(I + mR₂²)

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Answer:

Scientists have studied eclipses since ancient times. Aristotle observed that the Earth's shadow has a circular shape as it moves across the moon. He posited that this must mean the Earth was round. Another Greek astronomer named Aristarchus used a lunar eclipse to estimate the distance of the Moon and Sun from Earth

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If a coin has a mass of 29.34g and drops from a height of 14.7 meters, what is its loss in gravitational potential energy? Show
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Which statement supports Newton’s first law of motion?
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A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the
ddd [48]

A cart is pulled by horizontal force such that it moves with constant velocity

So here since velocity is constant we can say that its acceleration will be ZERO

now here for zero acceleration we can say

F_{net} = 0

So here we will have

F_{net} = F_{ap} - F_f = 0

here we know that

F_f = 10 N

so we will have

F_{ap} - 10 = 0

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so here applied force on handle will be

<em>b. 10 N</em>

5 0
3 years ago
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho
cupoosta [38]

Answer:

xf = 5.68 × 10³ m  

yf = 8.57 × 10³ m  

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

ax = 0                          ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t²     ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

yf = 8.57 × 10³ m  

5 0
3 years ago
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