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Makovka662 [10]

2 years ago

11

a pistol fires a bullet towards a target located 175m away. The bullet is traveling at 320 m/s. How long does it take for the bu

llet to hit the target?

1 answer:

inn [45]2 years ago

7 0

Time = distance/speed

Time = (175 m) / (320 m/s)

<em>Time = 0.547 second</em>

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Flura [38]

Answer:

about 4 km

Explanation:

15 minutes is a quarter of an hour, so you divide 16km by 4 to get your answer

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2 years ago

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp

sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

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3 years ago

A dog runs from point O to B. The dog reaches A at 2.3 s into the sprint. Then the dog reaches C at 4.1 s. What is the average v

anygoal [31]

It is 3.2 because you add 2.3 and 4.1 then divide your answer by 2.

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2 years ago

The level of mercury falls in a barometer<br>while taking it to a mountain.

Alexxandr [17]

Answer:

This is due to a relative decrease in atmospheric pressure in high places.

Explanation:

Given that atmospheric pressure decreases at the higher point or ground, this reduced atmospheric pressure, however, will be unable to contain the Mercury in the barometer tube.

Therefore, at the top of the mountain where the air pressure is low, the barometer reading ultimately goes down.

Hence, the level of mercury falls in a barometer while taking it to a mountain "due to a relative decrease in atmospheric pressure in high places."

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2 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

2 years ago