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sasho [114]
4 years ago
12

Two footballs (A & B) are thrown with the same amount of force. Football A has a greater mass. Describe the acceleration for

each football.
Physics
1 answer:
Sauron [17]4 years ago
8 0
Newton's second law gives the relationship between force applied to an object, its mass and its acceleration:
F=ma
where F is the force, m the mass and a the acceleration.

A force F is applied to football A, whose mass is m_A, and so the acceleration of this football will be given by (re-arranging the previous equation)
a_A =  \frac{F}{m_A}

Similarly, the acceleration of football B will be
a_B= \frac{F}{m_B}
where m_B is the mass of football B, and where the force F applied to the two footballs is the same.

Since football A has greater mass than football B, m_A \ \textgreater \  m_B, if we compare the two previous formula we see that the acceleration of football B is greater than the acceleration of football A:
a_B \ \textgreater \  a_A
Therefore, if the same force is applied to the two footballs, football B will accelerate more than football A.
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A 75.0 kg students sits 1.15 m away from a 68.4 kg student. What is the force of gravitational attraction between them?
AVprozaik [17]

Answer:

F=2.589×10⁻⁷ Newtons

Explanation:

The gravitational force of attraction between two bodies is given by

F= Gm₁m₂/r²

Where m₁ and m₂ are the masses of the two bodies, G is the universal gravitational constant while r is the distance between the two bodies.

G=6.67408 × 10⁻¹¹m³kg⁻¹s⁻²

r=1.15m

m₁=75.0kg

m₂=68.4kg

Therefore F= (6.67408×10⁻¹¹m³kg⁻¹s⁻²×75.0kg×68.4kg)/(1.15m)²

F=2.589×10⁻⁷ Newtons

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4 years ago
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3 years ago
Question # 40
jenyasd209 [6]

Answer:

Potential Energy = x = m g h

Kinetic energy = 1/2 m v^2

Assuming the mass fall from rest

1/2 m v^2 = m g h

v^2 = 2 g h

So the speed attained is independent of the mass

Also, x / v   does not have the units of mass

So the solution is none of the above.

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3 years ago
A penny is dropped from the Statue of Liberty
Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

vf=vi+at,

vf =unknown velocity  vi= initial velocity vi=0m/s  a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

Step 2: Solve

vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

x=vit+\frac{1}{2} at^{2}

Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0
3 years ago
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