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Romashka [77]
3 years ago
5

6. A yo-yo with a mass of 0.5 kg is tied to a radius of 2 m and is spinning at 3 m/s. What

Physics
1 answer:
pshichka [43]3 years ago
6 0

Given that,

Mass of yo-yo, m = 0.5 kg

Radius of the path, r = 2 m

Speed of the object, v = 3 m/s

To find,

The centripetal force acting on the yo-yo.

Solution,

The formula for the centripetal force is given as follows :

F=\dfrac{mv^2}{r}\\\\F=\dfrac{0.5\times (3)^2}{2}\\\\F=2.25\ N

So, the magnitude of centripetal force acting on the y-yo is 2.25 N.

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san4es73 [151]

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

dq = ρ(r)dV

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dV/dr = (4/3)×3πr^2

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dq = ρ(r)dV ; dq =ρ(r)4πr^2dr

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∫4Cπdr = 4Cπr

Q = 4Cπ(R - 0)

Q = 4CπR - 0

Q = 4CπR

C = Q/4πR

The value of C in terms of Q and R is [Q/4πR]

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3 years ago
A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be
nirvana33 [79]

Answer:

Work done, W = 19.6 J

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Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

W=\mu_kmgd

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So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

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