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arsen [322]
3 years ago
13

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull

et while it is traveling down the barrel of the rifle, 0.820 m long? assume constant acceleration and negligible friction.
Physics
1 answer:
muminat3 years ago
3 0

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.  

KE = 1/2*m*v^2

      = 1/2*0.005*320^2

      = 256 Joules.  

Divide this by the distance over which this energy was received and you have the force that provided that energy.  

  = 256/0.820 = 312.195 Newtons  

Rounded off, this is 312 N

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Explanation:

The temperature must be hot enough to allow the ions of deuterium and tritium to have enough kinetic energy to overcome the Coulomb barrier and fuse together. The ions must be confined with a high ion density to achieve a suitable fusion reaction rate.

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a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

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t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

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3 years ago
You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
dem82 [27]
  • Angle (θ) = 60°
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  • = (20 × 100) J
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<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

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Doubt clarification - use comment section.

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Answer:

Tp/Te = 2

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The orbital period of a planet around a star can be expressed mathematically as;

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Substituting the given values;

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So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

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