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Cerrena [4.2K]
3 years ago
8

the force of attraction between two masses is 3.60 Newtons if one of the masses is tripled and the distance between the masses i

s quadrupled what is the new force of attraction​
Physics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer: F = 19.2 N

Explanation: Given that the initial Force = 3.6N

The formula involved is

F = GMm/r^2

Substitute the force F

3.6 = GMm/r^2

If one of the masses is tripled and the distance between the masses is quadrupled. We have

3.6 = (G × 3Mm)/(4r)^2

Where G will be constant.

3.6 = 3GMm/16r^2

Separate the fraction of number

3.6 = 3/16 × GMm/r^2

Make GMm/r^2 the subject of formula

(3.6 × 16)/3 = GMm/r^2

19.2 = GMm/r^2

Therefore, the new force of attraction is 19.2 N

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What is the power of a circuit with voltage 12 V and current 8.0<br> A?
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a bowling ball has a mass of 6 kilograms. a tennis ball has a mass of 0.06 kilogram. how much inertia does the bowling ball have
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Answer:

100 times

Explanation:

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3 years ago
M=20kg F=10N d=25m<br>Work done=?​
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w=f×d

w=10N×25m

w=250Nm

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2 years ago
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia
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Answer with Explanation:

We are given that

m_1=0.3 kg

m_2=0.4 kg

Spring constant,k=200 N/m

h=0.250 m

a.Speed of ham before collision

u=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.25}=2.2m/s

Collision is inelastic

According to law of conservation of momentum

m_1u_1=(m_1+m_2)v

0.3\times 2.2=(0.3+0.4)v

v=\frac{0.3\times 2.2}{0.7}=0.94 m/s

Kinetic energy of masses=Spring potential energy at maximum distance

\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kA^2

A^2=\frac{(m_1+m_2)v^2}{k}

A=\sqrt{\frac{(m_1+m_2)v^2}{k}}

Substitute the values

A=\sqrt{\frac{(0.3+0.4)\times (0.94)^2}{200}}=0.056 m=0.056\times 100=5.6 cm

1 m=100 cm

b.Time period,T=2\pi\sqrt{\frac{(m_1+m_2)}{k}}

T=2\pi\sqrt{\frac{(0.3+0.4)}{200}}=0.37 s

7 0
3 years ago
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