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Phantasy [73]
3 years ago
15

A runner starts from a rest and speeds up with a constant acceleration . if she has gone a distance of 30 m at the point when sh

e reaches a speed of 8 m/s what is her acceleration
Physics
2 answers:
garik1379 [7]3 years ago
6 0

Answer:

a = 1.07 m/s^2

Explanation:

As we know that the runner starts from rest position

So here we can say

v_i = 0

now the distance moved by the runner is given as

d = 30 m

also we know that final speed of the runner will be

v_f = 8 m/s

now we know that

v_f^2 - v_i^2 = 2 a d

8^2 - 0 = 2(a)(30)

a = \frac{64}{60}

a = 1.07 m/s^2

arsen [322]3 years ago
5 0

Answer:

1.1 m/s

Explanation:

Apex

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When light passes through sequential interfaces, there may be a change of phase upon reflection at each interface.
Lena [83]

That's true.

I have a hunch that there definitely IS a change of phase at every reflection.

4 0
2 years ago
You weigh 580 N on Earth. If you were to go to Mars, where its gravitational pull is 3 . 7 11 m /s 2 , what would you weigh? (Hi
andrew-mc [135]

Answer:

59.18 kg

Explanation:

use f=ma

f= 580 N

a = 9.8 m/s 2

weigh(m) doesn't change only force(F) changes

7 0
3 years ago
Read 2 more answers
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
3 years ago
WOULUJUTUL RECIPECUIUS.
3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider M_{1} and M_{2} as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

                      \text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}

As gravitational constant G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}, M_{1} = 20 kg and  M_{2} = 100 kg, while d = 2.6 m, then

                    \text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}

Thus, we get finally,

                   \text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}

As we know, nano denoted by letter 'n' equals to 10^{-9}

So the force acting between two objects is 19.73 nN.

7 0
3 years ago
What does weight require?
kherson [118]

Answer:

kg

Explanation:

easy question to be honest

7 0
3 years ago
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