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FinnZ [79.3K]
3 years ago
11

A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?

Physics
1 answer:
stepladder [879]3 years ago
3 0
The 'strength' of the electric field is the force on 1C of charge at that point.

At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
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Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle \theta =51^{\circ}

launch velocity u=18 m/s

center of mass continue to travel its original Path so it center of mass will be at a distance of

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R=\frac{18^2\sin 102}{9.8}

R=32.33 m

Center of mass will be at x=32.33 m

(b)if one of the piece will be at x=26  m then other will be at

x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}

32.33=\frac{26+x_0}{2}

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8 0
3 years ago
A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight
photoshop1234 [79]

Answer:

t=1.9 sec

Explanation:

From the question we are told that:

Height h=28m

Time t=3s

Generally the Newton's equation for Initial velocity upward is mathematically given by

 s=ut+\frtac{1}{2}at^2

 28=3u-\frac{1}{2}*9.8*3^2

 u=24.03m/s

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 v=\sqrt{24.03-2*9.8*28}

 v=5.35m/s and -5.35m/s

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

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 5.35=24.03-9.8t

 t=\frac{28.03-5.35}{9.8}

 t=1.9 sec

4 0
2 years ago
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Rasek [7]

Answer:

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Explanation:

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5 0
3 years ago
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Hoochie [10]

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Explanation:

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3 years ago
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