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1 year ago

What conclusions can you make if a hockey goalie fails to block pucks shot in the upper right hand corner of the net

1 answer:

3 0

Given what we know, we can confirm that this result from the goalie is a clear indicator of room for improvement in the reaction speed and visual coordination for this area of the net.

<h3>How can the goalie improve reaction speeds to this area?</h3>

The key for situations like this is simply repetition. The more the goalie is able to practice with shots in this area of the net, the more muscle memory they will build regarding reacting to these shots, and therefore less time will be needed to block them in the future.

Therefore, we can confirm that this result from the goalie is a clear indicator of room for improvement in the reaction speed and visual coordination for this area of the net.

To learn more about reaction speeds visit:
brainly.com/question/8186329?referrer=searchResults

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What happens in the process of gravitational condensation?

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Answer:

An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.

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2 years ago

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ycow [4]

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2 years ago

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Explanation:

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2 years ago

1. What are the units of : pressure, force , area

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Pressure- Pascal( $N/m^2$ )

Force- Newton

Area- $m^2$ (Square metre)

<u>Explanation:</u>

Pressure is force continuous force applied to an object. The applied force is always perpendicular to the surface of the object.
Force is a push or pulls on an object by an external body where it changes its state from rest to motion.
Force is said to have both magnitude and direction.
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2 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

2 years ago