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stealth61 [152]

3 years ago

15

A 65.0 kg football player is running at 10.0 m/s downfield when he catches a 0.420 kg football thrown in the same direction but

at 31.0 m/s. How fast will the player and ball move after the catch

1 answer:

lukranit [14]3 years ago

8 0

Explanation:

Below is an attachment containing the solution.

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B is the right answer of the following statement

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An instrument that measures and detects vibrations in earth is known as a _______.

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A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound

Strike441 [17]

Answer:

$5.997m/s$

Explanation:

We were told to calculate the speed of the ball,

Given speed of sound as 340 m

And we know that the sound of the ball hitting the pins is at 2.80 s after the ball is released from his hands.

Speed of ball = distance traveled/(time of hearing - time the sound travels).

Speed= S/t

Where S= distance traveled

t= time of hearing - time the sound travels

time=time for ball to roll+timefor sound to come back.

time of sound=16.5/340

=0.048529secs

solving for speedof ball

Then,Speed of ball = distance traveled/(time of hearing - time the sound travels).

=16.5/(2.80-0.048529) m/s = 5.997m/s

Therefore, the speed of the ball is

5.997m/s

4 0

3 years ago

Which describes how a simple machine can make work easier ?

sveta [45]

A simple machine can make work easier by reduce the amount of energy needed to perform a task, therefore, B. <span>it magnifies the potential energy so that the kinetic energy is greater</span> is the correct answer.

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3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago