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Elan Coil [88]
2 years ago
11

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

uld the car have to accelerate so that a quarter placed on the back wall would remain in place?
Physics
1 answer:
djverab [1.8K]2 years ago
3 0

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

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8 0
3 years ago
What is the danger to spacecraft and astronauts from micrometeoroids?
Sloan [31]

Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)

Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s

KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules

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7 0
2 years ago
The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci
lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor =  2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

\dfrac{dE}{dt}=200 W

E =200\times 2 \times 10^{-6}

E =400 \times 10^{-6}\ J

we know,

E = \dfrac{1}{2}CV^2

V =\sqrt{\dfrac{2E}{C}}

V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}

V =\sqrt{400}

V = 20 V

so , the capacitor voltage is V = 20 V

7 0
3 years ago
A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s
djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, u=32 m/s

Acceleration of the driver is, a=-1.5 m/s²

Time taken to reach final velocity is, t=10.8 s.

The final velocity is given using the Newton's equations of motion as:

v=u+at, where, v is the final velocity.

Now, plug in the given values and solve for v.

v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}

Therefore, his final velocity is 15.8 m/s.

5 0
2 years ago
Explain how this happens?
grandymaker [24]

Answer:

When the polythene rod is rubbed with the woolen cloth, static electric charges move from the cloth and into the rod. The rod becomes negatively charged as negative charges move from the cloth and into the rod leaving the cloth positively charged as well.

5 0
3 years ago
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