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Nuetrik [128]
3 years ago
8

A hockey player swings her hockey stick and strikes a puck. According to Newtons 3rd law of motion which of the following is a r

eaction to the stick pushing on the puck
Physics
1 answer:
Korvikt [17]3 years ago
3 0

<u>Complete Question:</u>

A hockey player swings her hockey stick and strikes a puck. According to Newton’s third law of motion, which of the following is a reaction to the stick pushing on the puck?

A. the puck pushing on the stick .

B. the stick pushing on the player .

C. the player pushing on the stick .

D. the puck pushing on the player.

<u>Correct Option:</u>

According to Newton’s third law of motion the puck pushing on the stick is a reaction to the stick pushing on the puck.

<u>Option: A</u>

<u>Explanation:</u>

As when the hockey exert force on the puck (which is a flat ball basically used in ice hockey) then this action by hockey will receive equal and opposite reaction by puck. Thus when the stick is pushing on the this flat ball, then puck also push the stick. This is understood by newton's third law pf motion, where action and reaction forces are subject of discussion, displaying their is pair of forces applied among the interacting objects. This form is observed more practically in life and very frequent.

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Answer:

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a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what
yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

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4 0
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.

4 0
3 years ago
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