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docker41 [41]
3 years ago
8

Describe two examples of activities that you do where you depend on friction, and would not be able to do them without friction.

Physics
1 answer:
kaheart [24]3 years ago
7 0

Answer: One is when a bike stops. The other is when you rub your hands together to get warm.

Explanation: Force and friction affect our daily lives in numerous amounts of ways. For instance, when a football is kicked, it moves faster later after some time its force decreases due to friction. A common example of friciton is when a bike stops. When the brakes are applied the friction on the pads cause the bike to stop. The other is because you are getting warm with friction.

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Answer:

7. Your answer is correct dear, just add the unit

8. answer is 1.17m/s²

Explanation:

queation 7.

m = 3kg, F = 9N, a ?

F = ma

a = F/m = 9/3 = 3m/s²

Use the same approach for question 8

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The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a
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Answer:

19.2*10^6 s

Explanation:

The equation for time dilation is:

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}

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t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.

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Read 2 more answers
A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho
Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f

There is no kinetic energy in the initial state, nor potential energy in the end,

mgh+0=0+KE_f

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)

The inertia of the bodies is given by the equation,

I=\frac{m(R_1^2+R^2_2)}{2}

I=\frac{2(0.2^2+0.1^2)}{2}

I=0.05Kgm^2

On the other hand the angular velocity is given by

\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s

Replacing these values in the equation,

(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2

Solving for h,

h=2.86m

5 0
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