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3 years ago

What is an example of static friction

Physics

1 answer:

dolphi86 [110]3 years ago

7 0

imagine something setting stil, such as a box, since it is not moving, it has static friction

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For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a

Phantasy [73]

Torque is equal position vector times (r) times force vector (F). Since F= 10 N and r = 0.1 m, so the torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would be into the screen, clockwise rotation.

8 0

3 years ago

A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass

stealth61 [152]

Answer:

The shortest distance is $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

The distance between the axial is $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$ behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

For the bicycle not to be thrown over the

Momentum about the back wheel must be zero so

$\sum _B = 0$

=> $mg (26) = ma(40)$

=> $a = \frac{26}{40} g$

Here $g = 32.2 ft/s^2$

So $a = \frac{26}{40} (32.2)$

$a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

$v^2 = u^2 + 2as$

Where $u = 32.26 ft /s$

and $v = 0$ since the bicycle is coming to a stop

$v^2 = (32.26)^2 - 2(20.93) S$

=> $S = 24.86 ft$

5 0

3 years ago

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

mrs_skeptik [129]

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

$T - mgcos\theta = m\omega^2 R$

now we have

$T = mgcos\theta + m\omega^2 R$

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

$\omega = \frac{2\pi}{T}$

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

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3 years ago

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AlexFokin [52]

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