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melamori03 [73]
3 years ago
9

a solid sphere and a hollow sphere with equal mass are rotated about an axis through their centers. both spheres experience equa

l torque. which sphere will reach a speed of 10 rad/s first? Explain your answer.​
Physics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

Solid sphere

Explanation:

The solid sphere will have a lower moment of inertia as it carries more mass closer to the axis of rotation.

A lower moment of inertia results in a higher angular acceleration under a given torque.

α = τ/I

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Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other
denpristay [2]

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

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A 1.5m wire carries a 4 A current when a potential difference of 66 V is applied. What is the resistance of the wire?
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Example of the center of the gravity<br>​
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Explanation:

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3 years ago
I’ve already figured out A. I just need help with B and C.
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A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly
Ratling [72]

Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

v = 8.854 \ m/s

The tension in the string is:

T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N

Thus, the   tension in the string an instant before it broke = 34 N

6 0
3 years ago
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