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qaws [65]
3 years ago
15

Impulse is directly proportinal to both _____ and ____: greater is the force and/or time, greater is the ______​

Physics
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

to both 1)mass and 2)velocity, 3)momentum

Explanation:

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(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
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On which date is the gravitational force between Earth and the moon the greatest?
Genrish500 [490]

The gravitational forces between the Earth and Moon are greatest when the two bodies are closest together. That happens every 27.32 days, when the Moon is at the perigee of its orbit.

Even if this happened at the same time in every orbit, the date would change, because there are not 27.32 days in a month.

But it doesn't happen at the same time in every orbit ... the Moon's perigee precesses around its orbit, on account of the gravitational forces toward the Earth, the Sun, Venus, Mars, and the other planets.

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3 years ago
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.

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