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Illusion [34]
3 years ago
9

Buoyancy is a force that always acts in an

Physics
2 answers:
Maslowich3 years ago
8 0

Answer:

Explanation:

Buoyancy is a force that always acts in the upward direction against the weight of the immersed or semi-immersed object in the fluid. In any fluid( air or water) column pressure increases as we go deeper because of the overlying fluid. Here magnitude of the buoyancy force is proportional to the pressure difference.

Margarita [4]3 years ago
7 0

Buoyancy is a force that always acts in an upward direction exerted by a fluid on a body placed in the fluid

Hope this helps :)

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The answer is D correct me if I’m wrong
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This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?
Crazy boy [7]
1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction

2) F=ma
= (5.5kg)(30m/s)
=165 N
7 0
2 years ago
How would the moon appear from Earth if the moon did not rotate?
andrey2020 [161]
It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon
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3 years ago
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Compare the catching of two different water balloons.
Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B:  Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = F_{average} × Δt = mΔV

∴ F_{average} = m·ΔV/Δt

∴ For Case A F_{average} = 149.55×8/Δt =  1196.4/Δt N

For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

4 0
3 years ago
A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.​
GenaCL600 [577]

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

Force = 20N

Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

Impulse = 80Ns

Therefore, the impulse experienced by the object is 80 Newton-seconds.

7 0
2 years ago
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