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3 years ago

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

Physics

2 answers:

wlad13 [49]3 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]3 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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Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and

STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

here we know that

$\frac{I_1}{I_2} = 36 times$

also we know that

$r_1 = 10 Ly$

now we will have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$36 = \frac{r_2^2}{10^2}$

$r_2 = 60 Ly$

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3 years ago

4025.05m +20.0m +0.050004m

PIT_PIT [208]

The answer is 4,045.1 meters

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3 years ago

14. Which is the relationship between photon energy and frequency?

yawa3891 [41]

Answer:

The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy.

i hope this helps.

Explanation:

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2 years ago

In order to design an experiment, you need a ____ about the scientific question you are trying to answer.

goldenfox [79]

In order to design an experiment, you need a hypothesis about the scientific question you are trying to answer.

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3 years ago

Read 2 more answers

Waves in a fish bowl jostled by the Thingamajigger move to the sides at an average velocity of 0.50 m/s. If they occur once ever

Colt1911 [192]

Answer:

0.125 m

Explanation:

In this problem, we have:

v = 0.50 m/s is the average velocity of the wave

T = 0.25 s is the period of the wave

We can find the frequency of the wave, which is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.25 s}=4 Hz$

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

$\lambda=\frac{v}{f}$

Substituting the numerical values, we find

$\lambda=\frac{0.5 m/s}{4 Hz}=0.125 m$

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3 years ago