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Natalija [7]
3 years ago
11

A person holds a rifle horizontally and fires at a target. The bullet leaves the muzzle of the rifle with a velocity of 460 m/s.

Two seconds later the person hears the bullet strike the target.
If the speed of sound in air is 340 m/s, calculate the distance travelled from the bullet to the target.
Physics
1 answer:
Trava [24]3 years ago
3 0

Answer:

the distance travelled from the bullet to the target  is 391m

Explanation:

Hello! To solve this exercise we must follow the following steps.

1. the bullet travels with constant speed which means that the distance traveled to the target is given by the following equation

X=(V1)(T1)

T1=\frac{X}{V1} =\frac{x}{460}

where

X=target distance

V1=bullet speed=460m/s

T1=

time it takes for the bullet to reach the target

2. The distance the sound travels is given by the following equation (it is the same as the distance from the person to the target)

X=(V2)(T2)

T2=\frac{X}{V2} =\frac{x}{340}

X=

target distance

V2= speed of sound=340m/s

T2=   time it takes the sound of the Bullet to return.

3. The total time it takes for the person to hear the bullet(T=2s) is the sum of the time it takes for the bullet to reach the target, plus the time it takes for the sound to reach the person, with the above we infer the following equation

T=T1+T2

2=T1+T2

4. Finally we use the equations found in step 1 and 2 to find the distance traveled using algebra.

2=\frac{x}{340}+\frac{x}{460} \\x(\frac{1}{340} +\frac{1}{460} )=2\\\ X= \frac{2}{(\frac{1}{340} +\frac{1}{460} )} \\\\x=391m

the distance travelled from the bullet to the target  is 391m

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By taking into account that T1=3T2, we have

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3 years ago
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Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

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\boxed{ v= 15.9m/s}

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