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svetoff [14.1K]
3 years ago
14

PLS HELP Y’ALL I forgot how to do this :’(

Physics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

<em>Part A = </em><em>(  -7.072 , 7.072)</em>

<em />

<em>Part B = </em><em>( -5 , 0)</em>

Explanation:

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A motorcycle drives up a steeply inclined ramp. The work done on the motorcycle by the Earth’s gravitational force is ____. 1.
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Answer:

be

Explanation:

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A weightlifter pulls on a 225 kg bar with 1600 N of force, but the bar does not move. How much work is performed?
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Not alot

I think its 24.5N

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A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of
Aleksandr [31]

Answer:

Part 1)

L_1 = 185.2 kg m^2/s^2

Part 2)

L_2 = 663.07 kg m^2/s^2

Part 3)

L = 663.07 kg m^2/s^2

Part 4)

\omega = 1.83 rad/s

Part 5)

F_c = 453.6 N

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

L_1 = I \omega

here we know that

I = 194 kg m^2

\omega = 1.47 rad/s^2

now we have

L_1 = 194 \times 1.47

L_1 = 185.2 kg m^2/s^2

Part b)

Angular momentum of the person is given as

L = mvR

so we have

m = 68 kg

v = 4.9 m/s

R = 1.99 m

so we have

L_2 = (68)(4.9)(1.99)

L_2 = 663.07 kg m^2/s^2

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

L = 663.07 kg m^2/s^2

Part 4)

By angular momentum conservation of the system we will have

L_1 + L_2 = (I_1 + I_2)\omega

185.2 + 663.07 = (194 + 68(1.99^2))\omega

848.27 = 463.28 \omega

\omega = 1.83 rad/s

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

F_c = m\omega^2 R

F_c = 68(1.83^2)(1.99)

F_c = 453.6 N

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3 years ago
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Answer:

437 J

Explanation:

Parameters given:

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P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

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