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dusya [7]
4 years ago
7

1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force

Physics
1 answer:
Arturiano [62]4 years ago
4 0

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

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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}\\\\

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\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
Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta
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3 years ago
How to calculate moments with 3 separate weights of different amounts at different points?
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