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Ann [662]
3 years ago
6

10. What are the only types of passes allowed in floor hockey?​

Physics
2 answers:
patriot [66]3 years ago
7 0

Answer:

Your ans

Explanation:

For floor hockey

weqwewe [10]3 years ago
3 0

Answer:

the push pass and the wrist shot i believe

Explanation:

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Type the correct answer in the box. Round your answer to the nearest whole number. Calculate the man’s mass. (Use PE = m × g × h
Blababa [14]

Answer:

56 kg

Explanation:

The change in potential energy of the man is given by:

\Delta U = mg \Delta h

where

m is the man's mass

g is the gravitational acceleration

\Delta h is the change in height of the man

In this problem, we have:

\Delta U=4620 J is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

\Delta h=8.4 m is the change in height

Re-arranging the equation and substituting the numbers, we find the mass:

m=\frac{\Delta U}{g\Delta h}=\frac{4620 J}{(9.8 m/s^2)(8.4 m)}=56 kg

6 0
3 years ago
Read 2 more answers
Refer to the image. Calculate the magnitude of change in velocity of the cue ball.
valkas [14]

Answer:

If you add the 2 velocity vectors you get the velocity of the third vector which is the (change in) velocity of the cue ball after striking the cushion

V * 2 * cos 45 = 4 * cos 45 = 2.83 m/s

6 0
2 years ago
Newtons first law stateS that object will move With a constant velocity if nothing acts on it. Does our every day experience con
bazaltina [42]

Answer:

hi

Explanation:

hi

4 0
3 years ago
A steam engine absorbs 1.98 × 105 j and expels 1.49 × 105 j in each cycle. assume that all of the remaining energy is used to do
VikaD [51]
B) The heat absorbed by the engine is Q_{in} = 1.98 \cdot 10^5 J while the heat expelled is Q_{out} = 1.49 \cdot 10^5 J, therefore the work done by the engine is the difference between the heat absorbed and the heat expelled:
W=Q_{in} - Q_{out} = 1.98 \cdot 10^5 J - 1.49 \cdot 10^5 J = 0.49 \cdot 10^5 J

a) The efficiency of the engine is the ratio between the work done by the engine and the heat absorbed, therefore:
\eta= \frac{W}{Q_{in}} = \frac{0.49 \cdot 10^5 J}{1.98 \cdot 10^5 J}=0.247 = 24.7\%
8 0
4 years ago
A boat moves at 10.8 m/s relative to the water. If the boat is in a river where the current is 2.00 m/s, how long does it take t
Ray Of Light [21]

Answer:

Option E is the correct answer.

Explanation:

Velocity of boat = 10.8 m/s

Velocity of river = 2 m/s

Relative velocity upstream = 10.8 - 2 = 8.8 m/s

            Displacement = Velocity x Time

            1100 = 8.8 x t₁

        Time in upstream, t₁ = 125 s  

Relative velocity downstream = 10.8 + 2 = 12.8 m/s

            Displacement = Velocity x Time

            1100 = 12.8 x t₂

        Time in downstream, t₂ = 86 s  

Total time = t₁ + t₂ = 125 + 86 = 211 s

Option E is the correct answer.           

8 0
4 years ago
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