need help with #2, #3 Please thank you model to gl ve it an acceleration of 5 m/
ID: 1574827 • Letter: N
Question
need help with #2, #3 Please thank you
model to gl ve it an acceleration of 5 m/s2 2. A rifle bullet of mass 12.0 g has a muzzle velocity of 750 m/s. What is the average force acting on the bullet when the rifle is fired, if the bullet is accelerated over the entire 1.00-m length of the rifle? A steady force of 70.0 N, exerted 43.5 degrees above the horizontal, acts on a 30.0-kg sled on level snow. How far will the sled move in 8.50s? (Neglect friction) 4. A 20.0-kg block slides down a smooth inclined plane. The plane is 10.0 m long and is inclined at an angle of 30.0 degrees with the horizontal. Find (a) the acceleration of the block, and (b) the velocity of the block at the bottom of the plane. A 10.0-kg package slides down an inclined mail chute 15.0 m long. The top of the chute is 6.00 m above the floor. What is the speed of the package at the bottom of the chute if (a) the chute is frictionless and (b) the coefficient of kinetic friction is 0.3? 5. If a 4.00-kg container has á velocity of 3.00 m/s after sliding down a 2.00-m plane inclined at an angle of 30.0 degrees, what is (a) the force of friction acting on the 6. container and (b) the coefficient of friction between the container and the plane?Explanation / Answer
m = 12 g = 0.012 Kg
initial veleocity of bullet u = 0
Final velocity of bullet v = 750 m.s
length travelled x = 1m
acceleration a
using third equation of motion
v2 = u2 +2ax
7502 = 0 + 2a(1)
562500 = 2a
a = 281250 m/s2
F = ma
F = 0.012 (281250)
F = 3375 N
3) initial velocity u = 0
time t = 8.5 s
force in the horizontal direction
F = 70 cos 43.5
F = 50.78
a = F/m = 50.78/ 30
a = 1.69 m/s2
using second equation of motion
S = ut + (1/2)at2
S = 0 + (1/2)(1.69)(8.5)2
S = 61.1 m
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