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irina [24]
3 years ago
8

A 2 kg ball is attached by a string to the top of a vertical pole and swings in a circle of radius 1.2 m. If the string makes an

d angle of 18° with the pole, how fast is the ball moving?
Physics
1 answer:
never [62]3 years ago
5 0

Answer:

v = 3.81 m/s

Explanation:

First, we will find out the radius (r)

sinΘ = r / L

or

r = LsinΘ

on substituting the values, we get

r = 1.2 × sin18°

or

r = 0.370 m

Now,

The tension (T) can be calculated as

ΣF(y) = 0

or

Tcos18° - mg  = 0

or

T = mg / cos18.8°

on substituting the values, we have

T = (2kg x 9.8m/s²) / cos18°

or

T = 20.60 N

Now, applying the equilibrium condition in horizontal direction. we have

\Sigma F_x = ma_c = Tsin\theta = 0

also, the centripetal acceleration a_c=\frac{v}{r}

on substituting the values, we get

m(v² / r) = Tsin18°

v = (rTsin18° / m)

v = [(1.2m)(20.60 × sin18°) / 2kg]

v = 3.81 m/s

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maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
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p_i=p_f
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\frac{1}{2} (m_1+m_2)v_i^2
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\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
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Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

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