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Iteru [2.4K]
3 years ago
7

Suppose the mass of Object 1 is greater than that of Object 2 and that both objects are at rest until an explosion occurs.

Physics
1 answer:
Alexandra [31]3 years ago
4 0

Answer:

The objects exert the same force on each other.

Explanation:

Let us suppose that the mass of object 1 is greater than that of object 2. Both objects are at rest until an explosion occurs. The first object exert an equal and opposite force on another object as per Newton's third law of motion. In this case, Newton's third law will be applicable. As a result, he objects exert the same force on each other. Hence, the correct option is (a).

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Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.
sp2606 [1]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

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3 years ago
3. The expression 0.62 x10^3 is equivalent to...
Korolek [52]

\\ \sf\longmapsto 0.62\times 10^3

\\ \sf\longmapsto 62\times 10^{-2}\times 10^3

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\\ \sf\longmapsto 62\times 10

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3 years ago
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A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

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You are riding on a Ferris wheel that is rotating with constant speed. The car in which you are riding always maintains its corr
nirvana33 [79]
It's going down.

Hope this helps you
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Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero
Tatiana [17]

Answer:

The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.

6 0
3 years ago
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