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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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How the forces makes effect on the state of motion of an object ? <br><br> Plz ans fast urgent
Ksju [112]

Answer:

1. It may change the direction of an object in motion.

2. It may cause change in velocity of an object in motion.

Explanation:

1.It may change the direction of an object in motion.

When an object is in motion,an applied force on that object may change its direction.

For example, a sailboat moving eastward, can suddenly change its direction by interaction of a storm wind blowing form the south.

2. It may cause change in velocity of an object in motion .

A force applied to an object in motion can increase or decrease its speed. When the force is applied to the object in motion in the direction of that object, its velocity may increase.

On the other hand, when the force is applied in the opposite direction to the object in motion, its velocity may reduce.

6 0
3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

<u>v0 = 10 m/s</u>

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

<u>y = 1.2 m</u>

The maximum height above the ground is 1.2 m.

4 0
3 years ago
In many cartoon shows, a character runs of a cliff, realizes his predicament and lets out a scream. He continues to scream as he
Rzqust [24]

Answer:

Increasing until terminal velocity is reached

Explanation:

Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.

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The answer is true. The table does show an object moving with changing speed.
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