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daser333 [38]
3 years ago
8

A 45 g bullet strikes and becomes embedded in a 1.55 kg block of wood placed on a horizontal surface just in front of the gun. I

f the coefficient of kinetic friction between the block and the surface is 0.28, and the impact drives the block a distance of 13.0 m before it comes to rest, what was the muzzle speed of the bullet in meters/second?
Physics
1 answer:
zimovet [89]3 years ago
6 0

Answer:

299.51 m/s

Explanation:

m = mass of the bullet = 45 g = 0.045 kg

M = mass of the block = 1.55 kg

v = muzzle speed of the bullet

V = speed of bullet-block combination after the collision

μ = Coefficient of friction between the block and the surface = 0.28

d = distance traveled by the block = 13 m

V' = final speed of the bullet-block combination = 0 m/s

acceleration of the bullet-block combination due to frictional force is given as

a = - μg

using the kinematics equation

V'² = V² + 2 a d

0² = V² + 2 (- μg) d

0 = V² - 2 (μg) d

0 = V² - 2 (0.28) (9.8) (13)

V = 8.45 m/s

Using conservation of momentum for collision between bullet and block

mv = (M + m) V

(0.045) v = (1.55 + 0.045) (8.45)

v = 299.51 m/s

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Answer:

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Explanation:

Since the problem data is not complete, the following information is entered:

A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the  rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated  mixture.

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m_{f}=\frac{V_{f} }{v_{f} }  \\m_{g}=\frac{V_{g} }{v_{g} }  \\where:\\V_{f}=volume of the fluid[m^3]\\v_{f}=specific volume of the fluid [m^3/kg]\\

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V_{f}=1/3*V_{total}  \\V_{total}=1.78[m^3]\\

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m_{f}=\frac{1/3*1.78}{0.001190}  \\m_{f}=498.6[kg]\\m_{g}=\frac{2/3*1.78}{0.08609}\\m_{g}=\ 13.78[kg]

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