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Serhud [2]
3 years ago
11

A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward

the other?
Physics
1 answer:
zimovet [89]3 years ago
4 0

The gravitational force between the two balls is 6.13\cdot 10^{-9} N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem,  we have

m_1 = 6.8 kg

m_2 = 7.4 kg

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Answer:

Light are required on every trailer

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When a mass of 5 kg hangs from a vertical wire of length 2 m, waves travel on this wire with a speed of 10 m/s. What will the sp
liubo4ka [24]

Answer:

New speed of the wave, v = 5.77 m/s                                                                                                                                    

Explanation:

It is given that,

Mass of the object, m = 5 kg

Length of the wire, L = 2 m

Speed of the wave, v = 10 m/s

The speed of wave in terms of tension is given by :

v=\sqrt{\dfrac{T}{\mu}}

10=\sqrt{\dfrac{T}{\mu}}

If the wire is replaced with one with the same length but three times as heavy, the mass per unit length is given by :

\mu'=\dfrac{3m}{L}

New speed of the wave is given by :

v'=\sqrt{\dfrac{T}{\mu'}}

v'=\dfrac{1}{\sqrt{3} }\sqrt{\dfrac{T}{\mu}}

v'=\dfrac{10}{\sqrt{3} }

v' = 5.77 m/s

So, the speed of the new wave is 5.77 m/s. Hence, this is the required solution.

4 0
4 years ago
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2. What is the final velocity of an object that is traveling at 5.0 m/s and accelerates at
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Answer:

22.74 m/s

Explanation:

u= 5.0

a= 2.4

t= 7.39

Therefore the final velocity can be calculated as follows

v = u + at

v= 5 + 2.4(7.39)

v= 5 + 17.736

v= 22.74 m/s

Hence the final velocity of the object is 22.74 m/s

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A car and a semi-truck are both traveling along a road. Based on Newton's 2nd law, what will happen if both cars apply the same
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3 years ago
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Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Expl
professor190 [17]

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

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