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kherson [118]
3 years ago
12

How much force is required to accelerate a 2 kg mass at 3 m/s^2?*

Physics
1 answer:
alexgriva [62]3 years ago
5 0

Answer: 6N

Explanation:

F = MA due to Newton's second law

M = 2

A = 3

plug in to get

F = 2 * 3 = 6N

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A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?
fenix001 [56]
F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
7 0
3 years ago
How are electromagnetic waves different than all other waves?
Tems11 [23]
They differ from each other<span> in wavelength. Wavelength is the distance between </span>one wave<span> crest to the next. </span>Waves<span> in the </span>electromagnetic<span> spectrum vary in size from very long radio </span>waves<span> the size of buildings, to very short gamma-rays smaller </span>than<span> the size of the nucleus of an atom.</span>
8 0
3 years ago
Read 2 more answers
A man does 500 j work pushing a car a distance of 2m how much force does he apply
olasank [31]

Answer: 250 N

Explanation:

Use equation for work

W=F*d

d=2m

W=500J

F=?

-----------------------

W=Fd

F=W/d

F=500J/2m

F=250N

6 0
3 years ago
Read 2 more answers
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t
elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

       Angular velocity (\omega) = 2.23 rps

     Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

             = 2.23 revolutions per second

             = 2.23 \times 2 \times 3.14 rad/s

             = 14.01 rad/s

Now, tangential speed will be calculated as follows.

  Tangential speed, v = R \times \omega

                               = 0.379 x 14.01

                               = 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

8 0
3 years ago
A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the ob
Wittaler [7]

Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}  

u = 132.6 cm    

So, the object is placed 132.6 cm in front of the mirror.

(b) Magnification of the  mirror, m=\dfrac{-v}{u}

m=\dfrac{-17}{132.6}

m = -0.128

Hence, this is the required solution.

7 0
3 years ago
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