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MrRa [10]
3 years ago
10

when arti kicks a football two forces interact, arti's foot exerts a force of 5N on the ball. what size force will be exerted on

arti's foot?
Physics
2 answers:
nekit [7.7K]3 years ago
5 0

Answer:

5N

Explanation:

From Newton's third law, we understood that to every action, there is an equal an opposite reaction.

Since Arti's foot exerts a force of 5N on the ball, then, an equal and opposite force of 5N will be exerted on Arti's foot as understood from Newton's third law.

Therefore, a force of 5N will be exerted on Arti's foot.

CaHeK987 [17]3 years ago
3 0

Answer:

F=MA

5N=M×10M/SQ.

1/2 KG. =MASS

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The left hemisphere of the brain controls the right side of the body. Please select the best answer from the choices provided T
olga nikolaevna [1]

Answer:

True

Explanation:

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2 years ago
A bowling ball is dropped off the top of the Eiffel Tower. If the Eiffel Tower is 300 meters
patriot [66]

Answer:

7.82 s

Explanation:

Given:

Δy = 300 m

v₀ = 0 m/s

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(300 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 7.82 s

5 0
3 years ago
The force between charged objects decreases when their separation ..... A)increase<br> B)decrease
weqwewe [10]
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6 0
3 years ago
Is an object moving in uniform circular motion accelerating?
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8 0
3 years ago
A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the
Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion  that relates the height as follow:

h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth g=9.81\frac{m}{s^{2}}.

t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s

8 0
3 years ago
Read 2 more answers
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