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3 years ago

Distinguishing between Speed and Velocity.

Physics

1 answer:

koban [17]3 years ago

8 0

Answer:

Speed: Distance per time, 400 km/h, and a scalar quantity.

Velocity: Displacement per time, 20 m/s south, and a vector quantity.

Explanation:

Hope this helps! Please mark as brainliest.

Thanks!

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3 years ago

A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.

DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

$a=g=9.8 m/s^2$ is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

$ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s$

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

$v^2-u^2 = 2as'$

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

$s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m$

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0

3 years ago

A stone is thrown vertically upwards with an initial velocity of 20m/sec. Find the maximum height ot reaches and the time taken

MAXImum [283]

Answer:

The height reached is 20m, The time taken to reach 20m is 2 seconds

Explanation:

Observing the equations of motion we can see that the following equation will be most helpful for this question.

$v^{2} = u^{2} + 2as$

We are given initial velocity, u

We know that the stone will stop at its maximum height, so final velocity, v

Acceleration, a

And we are looking for the displacement (height reached), s

Substitute the values we are given into the equation

$0^{2} = 20^{2} + 2(10)s$

Rearrange for s

$0^{2} -20^{2} =20s$

$-400=20s$

$\frac{-400}{20} =s$

s = -20 (The negative is just showing direction, it can be ignored for now)

The height reached is 20m

Use a different equation to find the time taken

$s = vt - \frac{1}{2} at^{2}$

Substitute in the values we have

$-20=(0)t - \frac{1}{2} (10)t^{2}$

Rearrange for t

$-20 =0 -5 t^{2}$

$\frac{-20}{-5} =t^{2}$

$4 = t^{2}$

t = 2s

The time taken to reach 20m is 2 seconds

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3 years ago