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inn [45]
3 years ago
10

A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due

to gravity of 1.6 m/s2, a bullet shot straight into the air would take how long to return to the moon's surface? On Earth, a standard .30 caliber bullet fired from a rifle pointed straight up would rise to an altitude of 9,000 feet (2,743.2 meters) in 18 seconds, and then would return to Earth in another 31 seconds. How much influence does the atmosphere have on a bullet fired from the surface of the Earth?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2  

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

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QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0
3 years ago
If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.​
Alex17521 [72]

Answer: 200m^2

Explanation:

160N=32000N/x

x*160N=32000

x=200m^2

4 0
1 year ago
I need help, ASAP i’m failing and i have no clue what’s going on in my AP physics class at all.
garri49 [273]
What’s the question or problem ?
6 0
3 years ago
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