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Kisachek [45]
2 years ago
12

The main difference between a chest and a bounce pass is what?

Physics
2 answers:
wlad13 [49]2 years ago
6 0

Answer:

The main difference is: the chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate.

The secondary main difference is the amount of power from the ball recived from the person reciving

snow_lady [41]2 years ago
4 0

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.

Explanation:

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A heavier car is always safer in a crash than a lighter car.
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4 0
3 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
3 years ago
A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?
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The initial kinetic energy of the car is
E_1 =  \frac{1}{2}mv_1^2 =  1.9 \cdot 10^3 J

Then, the velocity of the car is decreased by half: v_2 =  \frac{v_1}{2}
so, the new kinetic energy is
E_2 =  \frac{1}{2}mv_2 ^2 =  \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
E_2 =  \frac{1.9 \cdot 10^3 J}{4}=475 J
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3 years ago
Which region of a longitudinal wave is this?
ipn [44]
I think the correct answer is is D.
8 0
2 years ago
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