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d1i1m1o1n [39]
3 years ago
15

How much force is required to accelerate a 5 kg mass at 20 m/s^2

Physics
1 answer:
Studentka2010 [4]3 years ago
6 0

Hello!

\large\boxed{F = 100N}

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

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How are domains arranged in materials that are magnetic and in ones that are not
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In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another

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If it takes 50n of force to lift a 450n what is the ma of the machine
otez555 [7]

If the machine is 100% efficient, then its
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If the machine is less than 100% efficient,
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7 0
3 years ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
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Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
3 years ago
A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second, and the t
slega [8]

Answer:

The tension increases to four times its original value.

Explanation:

v = Velocity

r = Radius

m = Mass of stone

The centripetal force is

F_c=m\dfrac{v^2}{r}

The tension will balance the centripetal force

T=m\dfrac{v^2}{r}

\\\Rightarrow T\propto v^2

\dfrac{T_1}{T_2}=\dfrac{v_1^2}{v_2^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{v_1}{2^2v_1^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{1}{4}\\\Rightarrow T_2=4T_1

The new tension will be 4 times the old tension

5 0
2 years ago
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