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Olenka [21]
3 years ago
7

Fill in the correct answers. A big wheel with twice the circumference of a small wheel will rotate with ____ the force and ___ t

he speed of the small wheel.
A) half, twice
B) Three times, half
C) twice, half
D) half, three times
Physics
2 answers:
hichkok12 [17]3 years ago
4 0
I think it’s c

hope this helped
Lina20 [59]3 years ago
3 0

Answer:

C.

Explanation:

""""""" Twice,Half """"""

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The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls l
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Let a be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

1.3mg - T = 1.3ma

where T is the magnitude of tension in the pulley cord, and

• for the mass on the right,

T - mg = ma

Eliminate T to get

(1.3mg - T) + (T - mg) = 1.3ma + ma

0.3mg = 2.3ma

\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}


Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity v such that

v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

mg - T' = ma'

where T' and a' are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

T' - 0.8mg = 0.8ma'

Eliminate T'.

(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'

0.2mg = 1.8 ma'

\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}

Now, the right-hand mass has an initial upward velocity of v, but we're now treating down as the positive direction. As it returns to its starting position, its speed v' at that point is such that

{v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}

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1 year ago
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FromTheMoon [43]

Answer:

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