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azamat
4 years ago
8

How long does it take a racehorse to accelerate from rest to a speed of 15 m/s with an acceleration of 3.0 m/s2?

Physics
2 answers:
GuDViN [60]4 years ago
8 0
It would take 5.0 s for a racehorse to accelerate from rest
aleksandr82 [10.1K]4 years ago
4 0

It would take 5.0 s for a racehorse to accelerate from rest to a speed of 15 m/s with an acceleration of 3.0 m/s.

Take 15 / 3.0 to get your answer of D. 5.0 s

Hope that helps u out!

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Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then
Andreas93 [3]

Answer:

The displacement is  \Delta H =    -  1 \ m

The distance  is  D =  4 \  m

Explanation:

From  the question we are told that

    The height from which the ball is dropped is  h  =  1 \ m

    The height attained at  the first bounce is  h_1  = 0.8  \  m

    The height attained at  the second bounce is   h_2 = 0.5 \  m

    The height attained at  the third bounce is h_3 = 0.2 \  m

Note  : When calculating displacement we consider the direction of motion

Generally given that upward is positive  the total displacement of the ball is mathematically represented as

            \Delta H =  (0  -  h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)

Here the 0 show that there was no bounce back to the point where Billy released the ball  

           \Delta H =  (0  -  1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)

=>          \Delta H =    -  1 \ m

Generally the distance covered by the ball is mathematically represented as  

                D =  h +  2h_2 + 2h_3 + 2h_3

The 2 shows that the ball traveled the height two times

              D =  1 +  2* 0.8  + 2* 0.5 + 2* 0.2

=>           D =  4 \  m

     

5 0
3 years ago
Two strong magnets were brought close to each other. They were repelling each other. Explain what must have happened.​
r-ruslan [8.4K]

When two magnets are brought together, the opposite poles will attract one another, but the like poles will repel one another. This is similar to electric charges. Like charges repel, and unlike charges attract.

pls. mark brainliest am. dyning for it

8 0
3 years ago
Read 2 more answers
Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elem
Aleksandr-060686 [28]

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus,  atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

7 0
3 years ago
01
klasskru [66]

The statement which describes how a machine can help make work easier is that It can put out more force than the input force by decreasing the distance over which force is applied, therefore the correct option is option A.

<h3>What is work done?</h3>

The total amount of energy transferred when a force is applied to move an object through some distance

The work done is the multiplication of applied force with displacement.

Work Done = Force * Displacement

As work done depends both on the force as well as the displacement force and be reduced by reducing the displacement if the same amount of work is performed by the machine.

The correct answer is option A since the statement that describes how a machine might assist in making work easier says that it can put out greater force than the input force by reducing the distance over which force is exerted.

Learn more about work done from here

brainly.com/question/13662169

#SPJ1

5 0
2 years ago
A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w
steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

h = L - L\cos{31°} = L(1 - \cos{31°})

so its initial gravitational potential energy U_0 is

U = mgh = mgL(1 - \cos{31°})

\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})

\:\:\:\:\:=0.23\:\text{J}

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U

We know that the initial kinetic energy K_0, as well as its final gravitational potential energy U are zero so we can write the conservation law as

mgL(1 - \cos{31°}) = \frac{1}{2}mv^2

Note that the mass gets cancelled out and then we solve for the velocity v as

v = \sqrt{2gL(1 - \cos{31°})}

\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}

\:\:\:\:\:= 1.3\:\text{m/s}

5 0
3 years ago
Read 2 more answers
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