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OverLord2011 [107]
3 years ago
12

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall.

Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return
Physics
1 answer:
xxMikexx [17]3 years ago
3 0

Answer:

36s

Explanation:

Let the objects be A and B.

Let the initial velocity of A be U and the initial velocity of B be 3U

The height sustain by A will be;

The final velocity would be zero

V2 = U2-2gH

Hence

0^2= U2 -2gH

H = U^2/2g

Similarly for object B, the height sustain is;

V2 = (3U)^2-2gH

Hence

0^2= 3U^2 -2gH

U2-2gH

Hence

0^2= U2 -2gH

H = 3U^2/2g

By comparism. The object with higher velocity sustains more height and so should fall longer than object A.

Now object A would take;

From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;

V=10×12=120m/s let g be 10m/S2

Similarly for object B,

The final velocity for B when it's falling it should be 3×that of A

Meaning

3V= gt

t =3V/g = 3× 120/10 = 36s

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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
Tju [1.3M]

Answer:

675m

Explanation:

Given parameters:

Initial velocity  = 0m/s

Acceleration  = 6m/s²

Time  = 15s

Unknown:

Distance traveled by the body = ?

Solution:

To solve this problem; we use the expression;

   S = ut + \frac{1}{2}at²

 Where u is the initial velocity

              t is the time

             a is the temperature

         

Insert the parameters and solve;

  S = 0 x 15 + \frac{1}{2} x 6 x 15²  

  S = 675m

5 0
2 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 93 m long. It was a rainy day and the coeffi
USPshnik [31]

Answer:

u = 25 m/s

Explanation:

given,                                

length of skid = 93 m          

coefficient of friction = 0.35

final velocity = 0 m/s              

initial velocity = ?                        

force here is friction  f = μ mg

F = ma                                                

now com paring                      

-μ mg = m a                      

a = - μ g                    

a = - 0.35 x 9.8              

a = -3.43 m/s²

we know,              

v² = u² + 2 a s                        

0 = u² - 2 x 3.43 x 93                

u² = 637.98                    

u = 25.26 m/s                      

u = 25 m/s (two significant figure)

6 0
3 years ago
How should the student use the data collected and the known quantities from the experiment to determine the initial total mechan
fiasKO [112]

It should be noted that the student should use K = 1/2mv² with the initial speed of the block for one trial.

<h3>Method of using the data collected.</h3>

From the complete information, the student wants to use the data collected and the known quantities from the experiment to determine the initial total mechanical energy of the block-ramp-Earth system for all trials in the experiment.

In this case, it's important to use K = 1/2mv² with the block's initial speed for one trial due to the fact that the initial speed is the same in all the trials.

Learn more about experiments on:

brainly.com/question/17274244

7 0
2 years ago
Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What
IgorLugansk [536]

Answer:

Frequency is 213.04\ s^{-1}.

Explanation:

Distance between source 1 from the receiver , S_1 =\sqrt{10^2+22^2}=24.17\ m.

Distance between source 2 from the receiver , S_2=\sqrt{5^2+22^2}=22.56\ m.

Now ,

Path difference , r = S_1-S_2=24.17-22.56=1.61\ m.

We know, for constructive interference path difference should be integral multiple of wavelength .  

Therefore, r=n\times \lambda

It is given that n = 1,

Therefore, \lambda=1.61\ m.

Frequency can be found by , \nu=\dfrac{v}{\lambda}= \dfrac{343}{1.61}=   213.04\ s^{-1} .

Hence, this is the required solution.

5 0
3 years ago
1. What is the most common type of distribution? How does this distribution<br> benefit the species?
timurjin [86]

Answer:

Clumped distribution is the most common type of dispersion found in nature. In clumped distribution, the distance between neighboring individuals is minimized.

6 0
2 years ago
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