Respuesta :
Answer:
Explanation:
calculate the momentum before the puck hits the goalie: momentum = mass*velocity
p = 0.17*54.9
p = 9.333 kg.m/s
Because the final momentum is zero, the change in momentum is also 9.333, which is also called impulse.
Apply impulse = Force*time
9.333 = 3.62*10^-3*F
F = 2578.17679558
Therefore the average force exerted is 2578N
First few instances I performed goalie, I wore a participant helmet with cage + neck protector. (Cooper HH 3000 if it concerns) in case you could purchase some extra padding and function it placed into the goalie helmet. Use of a "doo-rag" (kinda feels like a yamulkha) will soak up a number of the surplus room as properly. the burden is greater on the grounds which you at the instant are not used to it. yet in the top, it is your protection on the line.
A hockey goalie is standing on ice. Another player fires a puck (m = 0.170 kg) at the goalie with a velocity of +44.6 m/s. The magnitude of the average force exerted on the goalie by the puck is 1490.04 N for (a) and 3020.72 N for (b)
From the information given:
- the initial speed (u) of the puck fired to the hockey goalie is = 44.6 m/s
- time (t) at which the goalie catches the puck = 5.02 x 10-3 s
The magnitude of the average force is the change in the momentum divided by time (t).
(a)
The magnitude of the average force is:
[tex]\mathbf{F_{avg}= \dfrac{0.170 \ kg \times 44.6 \ m/s }{5.02 \times 10^{-3} \ s }}[/tex]
[tex]\mathbf{=1490.04 \ N}[/tex]
(b)
Suppose;
- the final speed = -44.6 m/s
The magnitude of the average force can be computed as:
[tex]= \mathbf{\dfrac{0.170 \times (44.6 -(-44.6))}{5.02 \times 10^{-3}} }[/tex]
[tex]= \mathbf{\dfrac{0.170 \times (44.6 +44.6)}{5.02 \times 10^{-3}} }[/tex]
= 3020.72 N
Learn more about the magnitude of the average force here:
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