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irga5000 [103]
3 years ago
14

A rifle bullet with a mass of 11.5 g traveling toward the right at 251 m/s strikes a large bag of sand and penetrates it to a de

pth of 23.7 cm. determine the magnitude and direction of the friction force (assumed constant) that acts on the bullet.
Physics
1 answer:
BabaBlast [244]3 years ago
5 0

To look for the acceleration, it will come from:

vf^2=v0^2+2ad 
where:
vf = final velocity = 0 
v0 = initial velocity =251 m/s 
a = acceleration 
d= distance traveled = 0.237 m 

0=251^2+2a(0.237 ) 
a= -251 ^2 / (2*0.237) =-132 913.502 m/s/s 

we find the force from: 

F = ma = 0.0115kg*(-1.32x10^5m/s/s) = -1518 N 

the negative sign shows that the force is in the direction contradictory the bullet's motion

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The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, what will the pressure of a gas be after the temperature changes?

8 0
3 years ago
A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

so now the weight of the rifle=\frac{25}{10}  

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0
3 years ago
Una rueda que tiene 15 cm de radio, realiza 64 vueltas en 16 seg. Calcula: Periodo Frecuencia Velocidad angular Velocidad lineal
liraira [26]

Answer:

i) El período de la rueda es de 0,25 segundos.

ii) La frecuencia de la rueda es 4 Hertz

iii) La velocidad angular es aproximadamente 25.133

iv) La velocidad lineal es de aproximadamente 3,77 m / s

Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

7 0
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How much force is needed to accelerate a 20 kg mass at a rate of 4 m/s to the second power?
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Answer:

So 55 Newtons are needed.

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