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icang [17]
3 years ago
5

What is the difference between speed and velocity?

Physics
1 answer:
Wittaler [7]3 years ago
7 0
It’s C) because we know that velocity is a vector quantity with direction and speed is only quantity. For example, velocity could be 5 m/s to the east but speed would only be 5 m/s
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A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area
Airida [17]
For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
p_1 = p_2
which becomes
\frac{F_1}{A_1}= \frac{F_2}{A_2}
where
F_1=3600 N is the force on the first piston
A_1=0.030 m^2 is the area of the first piston
F_2 is the force on the second piston
A_2=0.015 m^2 is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
F_2=F_1  \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N
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3 years ago
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A musical tone sounded on a pi has frequency of 410 Hz and a wavelength of 0.80 m.
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C. I did it bit i forgot how I did it
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Before using wind energy to produce electricity, humans used wind energy to produce PLEASE ANSWER QUICKLY
topjm [15]

Answer:

Water

Explanation:

windmills were used to bring water to farms in ancient times.

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Which wave has a disturbance that is parallel to the wave motion?
Leto [7]
The answer is D.<span>longitudinal</span>
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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
2 years ago
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