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Lady_Fox [76]
2 years ago
10

1. How does the acceleration compare for an object in free fall when an it is thrown up versus thrown down? a.The acceleration i

s faster when it is thrown up compared to thrown down. b.The acceleration is positive when it is thrown up and negative when it is thrown down. c.The acceleration is slower when it is thrown up compared to thrown down. d.The acceleration is the same regardless of being thrown up or thrown down.
2. A textbook is dropped from the second story stairs and free falls to the ground. What changes, if any, would be observed of the velocity and the acceleration of the textbook as it falls?a.The velocity decreases b.The velocity increases c.The acceleration remains constantd. Both B & C
Physics
2 answers:
alexgriva [62]2 years ago
7 0

Answer:

1) Option b. The acceleration is positive when it is thrown up and negative when it is thrown down.

2) Option d. Both B & C

Explanation:

Question 1)

Acceleration is defined as the rate of change of velocity with respect to time.

a=\frac{v_{f}-v_{i}}{t}

When an object is thrown up, its velocity is maximum at the initial point and decreases as the object moves up. Since, the velocity is decreasing, the acceleration will be negative. This can also be proven from the formula mentioned above. When the object is thrown up, initial velocity(vi) is maximum and the final velocity (v_f) is zero, as the object rests for a very tiny moment at its maximum point. This will give a negative numerator and hence the value of acceleration will be negative.

When an object is thrown down, its velocity is zero at the start of the motion and increases as the object falls down. Since, the velocity is increasing in this case, the acceleration will be positive.

Therefore, Option b, gives the correct answer.

Question 2)

When an object falls freely, it falls under the action of gravity under a constant acceleration known as gravitation acceleration. Gravitational acceleration is represented by g and is equal to 9.8 m/s².

Since, the textbook is falling freely, its acceleration will be constant i.e. equal to gravitational acceleration.

When the object falls down, with every second its velocity increases. The velocity in minimum(zero) at the start and is maximum just before the object hits the ground.

So, when the textbook is dropped, it will fall with constant acceleration and its velocity will be increasing.

Therefore, for this question, Both B and C are correct.

Scrat [10]2 years ago
6 0

1.

Answer:

d.The acceleration is the same regardless of being thrown up or thrown down.

Explanation:

As we know that when object is in air then net force on object is given as

F = F_g

here we know that

F_g = mg

now we will have

ma = mg

so we can say

a = g

so in any case whenever object is in air its acceleration is vertically downwards and magnitude is equal to "g"

Answer:

Both B & C

Explanation:

As we know that when book is in free fall then due to weight of the book it will experience net downward force.

This force will exert

F = mg

now we know that

a = g

so while in free fall the acceleration due to gravity will remain constant on the book but speed will increase

so we have

b.The velocity increases

c.The acceleration remains constant

d. Both B & C

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So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

\boxed{\sf{\bold{t = \frac{s}{v}}}}

With the following condition :

  • t = interval of the time (s)
  • s = shift or displacement (m)
  • v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

  • h = height or any other displacement at vertical line = 19.6 m
  • g = acceleration of the gravity = 9.8 m/s²
  • v = velocity = 343 m/s

What was asked :

  • \sf{\sum t} = ... s

Step by step :

  • Find the time when the object falls freely until it hits the water. Save value as \sf{\bold{t_1}}

\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}

\sf{t_1 = \sqrt{4}}

\sf{\bold{t_1 = 2 \: s}}

  • Find the time when the sound propagate through air. Save value as \sf{\bold{t_2}}

\sf{t_2 = \frac{h}{v}}

\sf{t_2 = \frac{19.6}{343}}

\sf{\bold{t_2 \approx 0.06 \: s}}

  • Find the total time \sf{\bold{\sum t}}

\sf{\sum t = t_1 + t_2}

\sf{\sum t \approx 2 + 0.06}

\boxed{\sf{\sum t \approx 2.06}}

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

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