All categories

3 years ago

A bullet leaves a rifle with a muzzle velocity of 521 m/s. while accelerating through the barrel of the rifle, the bullet moves

a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration)

Physics

1 answer:

iVinArrow [24]3 years ago

6 0

$velocity=\frac{distance}{time}\\\\
velocity*time=distance\\\\
time=\frac{distance}{velocity}\\\\
time=\frac{0,840m}{521\frac{m}{s}}=0,0016s\\\\
acceleration=\frac{velocity}{time}=\frac{521}{0,0016}=325625\frac{m}{s^2}$

You might be interested in

If a canadian plane is traveling at 500 miles/hr, how fast is this in m/s?

patriot [66]

223.52 meters per second

6 0

3 years ago

a business that sells products to teens would most likely create a website with a title ending in what suffix?

Nataliya [291]

I think you forgot to give the options along with your question. I am answering the question based on my knowledge and research. A business that sells products to teens would most likely create a website with a title ending in .com. I hope that this is the answer that has actually come to your great help.

3 0

2 years ago

Read 2 more answers

What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.

soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0

2 years ago

4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat

nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0

3 years ago

12. A las 10 de la mañana Elena sale a 100 Km/h de una ciudad A con dirección a Madrid. A la misma hora sale Javier desde otra c

Klio2033 [76]

Answer:

a) t = 3.3 [h]

b ) Hora = 13:18 o 1:18 [pm]

c) x = 327.79 [km/h] (Elena)

x = 196.6 [km] (Javier)

Explanation:

Para poder solucionar este problema debemos hacer un planteamiento inicial de ubicacion de las ciudades, este planteamiento nos ayudara a entender el problema de una manera mas facil.

Tenemos las ciudades A & B y la ciudad de Madrid que esta a una distancia x con respecto de B, (ver esquema adjunto).

De manera logica debemos deducir que la Ciudad A debe estar mas lejos de Madrid que la ciudad B de la misma Madrid, ya que en caso contrario Javier nunca alcanzara a Elena, ya que Elena va mas rapido que Javier.

a) Ahora debemos de utilizar la siguiene ecuacion de la cinematica, cuando los cuerpos se mueven a velocidad constante.

$x=x_{o}+v*t$

Donde:

x -xo = Distancia entre el punto inicial y punto final.

v = velocidad [m/s]

t = tiempo [s]

Debemos convertir las velocidades de kilometros por hora a metros por segundo.

$100 [\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 27.77[m/s]\\60[\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 16.66[m/s]$

Seguidamente formulamos una ecuacion por cada movimiento, luego debemos igualar estas ecuaciones en funcion de la variable x que sera el punto donde se encuentren ambas personas.

Para Elena

$(132000+x) = 27.77*t\\x = 27.77*t - 132000$