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Marrrta [24]
3 years ago
14

The gravitational force between two objects is 3600 N. What will be the gravitational force between the objects if the mass of e

ach object is reduced to one third of its original mass
A. 400 N
B. 1200 N
C. 10,800 N
D. 32,400
Physics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

It will reduce by 9 times and become 400 N.

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When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
  • Gravitational attraction force is GMm/r² between the earth and a person.
  • M= mass of the earth

m= mass of the person

r= separation between them.

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Physics 1
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The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.

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The given data in the problem is;

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a = F/m

Explanation:

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A mouse ran 100 centimeters in
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Read 2 more answers
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load m_{LS} = 10000 kg

mass of flat bed m_{FB} = 20000 kg

initial speed of truck v_{0} = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs F_{N}     -------------let this be equation 1

where F_{N} = normal force = mg

so

Fs,max = μs mg

ma_{max} = μs mg

divide through by mass

a_{max} = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a_{max} = 0.5 × 9.8 m/s²

a_{max} = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

v_{f}² = v_{0}² + 2aΔx

where v_{f} is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0
3 years ago
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