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mr_godi [17]
3 years ago
11

A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du

e to air resistance increases. How does this affect the acceleration of the penny?
a. The acceleration decreases b. The acceleration remains constant and not zero (my answer)
c. The acceleration remains zero d. The acceleration increases.
Physics
2 answers:
pentagon [3]3 years ago
8 0
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
yuradex [85]3 years ago
3 0

Answer:

The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.

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Here when car in front of us applied brakes then it is slowing down due to frictional force on it

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F_f = \mu m g

So the acceleration of car due to friction is given as

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a = \frac{F_{net}}{m}

a = -\mu g

now it is given that

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so here we have

a = -0.868 * 9.81

a = -8.52 m/s^2

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0
3 years ago
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Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

Now, the kinetic energy of the particle as it moves through point B is:

\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

8 0
3 years ago
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111.333

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Answer:

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Answer:

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Because of the Coriolis effect, which deviates the atmosphere in the left-hand side of the southern hemisphere and generates a southeastern wind, this wind did not blow directly from south to east.

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