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2 years ago

What affects the way a projectile performs when it is shot from a firearm?

Physics

2 answers:

adoni [48]2 years ago

6 0

Answer:

In no particular order, they are:

Propellant: The amount of propellant and the type of propellant will determine the ultimate range of the projectile.

Muzzle Length: In accordance with propellant, determines the muzzle velocity of the projectile. A general rule of thumb is, the longer the muzzle, the higher the muzzle velocity.

Projectile Type: As in it's shape. A baseball is not going to fly as far as an artillery shell, even if thrown at the same speeds, due to the drag it creates.

Atmospheric Pressure: Affects how dense the air is, determines how much drag the projectile will have to fly through, affecting it's range.

Temperature: Same as atmospheric pressure.

Wind: Depending on the speed and direction, can result in the projectile arriving in places it has no business being in.

Humidity: Same as atmospheric pressure, but also if a projectile has to travel through a particularly humid area when the conditions are just right and if the projectile is in the air long enough, it can accumulate ice, affecting it's shape, increasing drag and increasing it's weight.

Spin of the Earth, or The Coriolis Effect: Shows it's effects most at the West-East, and vice versa, directions. Can be negligible in short distances, such as when a pistol is fired or a ball is thrown. Will become more noticeable as the range increases and will need corrections to aiming such as when firing a sniper rifle at long range or artillery fire. Projectiles fired towards the East will end up higher than the aim point, and those that're fired West will end up lower. In long enough ranges it can cause the projectile to miss a whole city.

Spin of the Projectile: Almost all long range artillery and firearms will induce a spin on the projectile through rifling in the gun barrel. Rifling and the rate of spin will depend on the particular gun and it's design/purpose. Spinning a projectile will make it more stable in flight, ergo making it more accurate. But that spin comes from redirecting some of the kinetic energy during propulsion and will cause the projectile to have less range compared to an identical smoothbore gun with the same propellant.

Gravity: Not a major consideration for most applications on Earth truth be told, but is crucial when said projectile is designed to go to space, like rockets.

These are what I could remember with sleepy eyes, having just woken up. It should be comprehensive enough, but if I remember one I missed, I'll be sure to add it here.

coldgirl [10]2 years ago

4 0

Answer:

the type and amount of gunpowder

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two negative charge of -2.0 c and a positive charge of 3.0 c are separated by 80 m what is the force between two charges

kozerog [31]

The Coulomb force is equal to the constant k times the product of charge one and charge two over radius.

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2 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

7 0

2 years ago

A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh

elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

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2 years ago

A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

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2 years ago

A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app

Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0

2 years ago