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Dominik [7]
3 years ago
13

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

tions every 10 seconds. what is the centripetal acceleration and velocity felt by the ball?
Physics
1 answer:
andrezito [222]3 years ago
4 0

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

f = \frac{5}{10} = 0.5 Hz

now we know that angular frequency is given as

\omega = 2\pi f

\omega = 2(\pi)(0.5)

\omega = \pi rad/s

Now centripetal acceleration is given as

a_c = \omega^2 R

a_c = \pi^2 (0.7)

a_c = 6.9 m/s^2

now the velocity of the ball at this angular frequency is given as

v = R\omega

v = 0.7 (\pi)

v = 2.2 m/s

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In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
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Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
3 years ago
Which of the following statements it's true about Sir isaac Newton
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3 years ago
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A charged particle is placed in an external magnetic field and it is moving in a circular path of radius 26m.The magnetic force
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Answer:

208 Joules

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Kinetic energy, K.E. = (1/2)·m·v²

Where;

m = The mass of the charged particle

v = The velocity of the charged particle

r = The radius of the path of the charged particle

Whereby the magnetic force acting on the charge particle = The centripetal force, we have;

F = F_c = m·v²/r = 16 N

(1/2) × r × F_c = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.

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∴ 208 Joules = K.E.

The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.

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3 years ago
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