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LuckyWell [14K]
2 years ago
11

An object is projected horizontally at 18.4 m/s from the top of a 162.7 meter cliff.

Physics
1 answer:
miskamm [114]2 years ago
5 0

Answer:

  106.03 meters  

Explanation:

The height is given by the formula for motion under the influence of gravity.

  h = -4.9t^2 +162.7

Height is 0 when ...

  0 = -4.9t^2 +162.7

  4.9t^2 = 162.7

  t^2 = 162.7/4.9

  t = √(162.7/4.9)

The horizontal distance traveled in that time is ...

  (18.4 m/s)√(162.7/4.9) s ≈ 106.03 m

The object will strike the ground about 106.03 meters from the base of the cliff.

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What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo
Andreyy89

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees  downward from the horizontal?

Answer:

W_{work}=2.67*10^{-3}J

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

W_{work}=F_{force}*D_{distance}Cos\alpha  \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J

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