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Gelneren [198K]
3 years ago
15

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?

Physics
1 answer:
AVprozaik [17]3 years ago
7 0

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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A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?
ycow [4]

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

= 27 joules

Thus, 27 joules of work is done on the bench.

6 0
3 years ago
If the only forces acting on a 2.0kg mass are F1 = (3i-8j)N and F2 = (5i+3j)N, what is the magnitude of the acceleration of the
belka [17]

Answer: 4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

3 0
3 years ago
Atoms in a molecule are bonded together by sharing gaining or losing_
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It is by gaining! I just learned this in school. Hope this helps:)

3 0
3 years ago
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A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang
erma4kov [3.2K]

Given:

initial angular speed, \omega _{i} = 21.5 rad/s

final angular speed, \omega _{f} = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

\alpha = \frac{\omega_{f} - \omega _{i}}{t}

Now, putting the given values in the above formula:

\alpha = \frac{28.0 - 21.5}{3.50}

\alpha = 1.86 m/s^{2}

Therefore, angular acceleration is:

\alpha = 1.86 m/s^{2}

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3 years ago
High energy waves have short wave lengths, true or false?
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True, the wavelength dies down due to high frequency and low amptitude. 
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