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IgorC [24]
2 years ago
11

Hollywood and video games often depict the bad guys being "blown away" when they’re shot by a bullet (i.e. once hit, their feet

leave the ground and they fly backwards). Assuming that even if a handgun cartridge did generate enough momentum for the bullet to do this, why is it still nonsense on-screen?
Physics
1 answer:
Anit [1.1K]2 years ago
4 0

Answer:

Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".

Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.  

What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.  

The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.  

It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this,  it is still nonsense on screen in Hollywood and video.  

             

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2 years ago
What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl
trasher [3.6K]

Net flux through the cylindrical surface is given as

\phi = \frac{q}{epsilon_0}

here q = enclosed charge in the surface

so here in order to find the value of q

q = \lambda* L

so now we have

\phi = \frac{\lambda * L}{\epsilon_0}

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now by Gauss's law we can find the electric field

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8 0
2 years ago
A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410,
Alik [6]
  • The horizontal pushing force required to just start the crate moving is 447 N.
  • The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

<u>Explanation</u>:

  • By the definition of the coefficient of static friction we have:

                                 μ_{s} = \frac{F_{appl} }{W}= \frac{F_{s} }{N},

where, F_{appl} is the horizontal pushing force,

            W = mg is the weight of the crate directed downward,

            F_{s} is  the static  friction  force-directed  opposite  to  the  horizontal  pushing force and equal to it,

            N is the force of reaction directed upward and equal to the weight of the crate.

From  this  formula  we  can  find the horizontal pushing  force  required to  just  start the crate moving:

                         F_{appl} = F_{s} = u_{s}N = u_{s}mg

                                                      = 0.760 \times 60 kg \times 9.8 m / s^2

                                                      = 447 N.

  • By the definition of the coefficient of kinetic friction we have:

                              u_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N},

where, F_{appl} is the horizontal pushing force,

            W = mg is the weight of the crate directed downward,

            F_{k} is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,

            N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:

                              F_{appl} = F_{k} = u_{k}N = u_{k}mg

                                                            = 0.410 \times 60 \times 9.8

                                                            = 241 N.

  • The horizontal pushing force required to just start the crate moving is 447 N.
  • The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.
6 0
2 years ago
A standing wave of 603 Hz is produced on a string that is 1.33 m long and fixed on both ends. If the speed of the waves on this
kondaur [170]

Answer:

4.

Explanation:

Given,

frequency of standing wave = 603 Hz

length of string,L = 1.33 m

speed of the wave, v = 402 m/s

number of antinodes = ?

Wavelength of the standing wave

\lambda = \dfrac{v}{f}

\lambda = \dfrac{402}{603}

\lambda = 0.67\ m

Number of anti nodes in the standing wave

n=\dfrac{l}{\frac{\lambda}{2}}

n=\dfrac{2l}{\lambda}

n=\dfrac{2\times 1.33}{0.67}

n =3.97= 4.

Number of antinodes is equal to 4.

7 0
2 years ago
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