Answer:
The distance from the top of the stick would be 2l/3
Explanation:
Let the impulse 'FΔt' acts as a distance 'x' from the hinge 'H'. Assume no impulsive reaction is generated at 'H'. Let the angular velocity of the rod about 'H' just after the applied impulse be 'W'. Also consider that the center of percussion is the point on a bean attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot.
Applying impulse momentum theorem for linear momentum.
FΔt = m(Wl/2), since velocity of center of mass of rod = Wl/2
Similarly applying impulse momentum theorem per angular momentum about H
FΔt * x = I * W
Where FΔt * x represents the impulsive torque and I is the moment of inertia
F Δt.x = (ml² . W)/3
Substituting FΔt
M(Wl/2) * x = (ml². W)/3
1/x = 3/2l
x = 2l/3
Answer:
(B) 1.6 m/s^2
Explanation:
The equation of the forces acting on the box in the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, 
being the angle of the incline
is the frictional force, with 
being the coefficient of kinetic friction, N being the normal reaction of the plane
a is the acceleration
The equation of the force along the direction perpendicular to the slope is
where 
is the component of the weight in the direction perpendicular to the slope. Solving for N,
Substituting into (1), solving for a, we find the acceleration:
Answer;
= 312 Newtons
Explanation;
The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.
KE = 1/2*m*v^2
= 1/2*0.005*320^2
= 256 Joules.
Divide this by the distance over which this energy was received and you have the force that provided that energy.
= 256/0.820 = 312.195 Newtons
Rounded off, this is 312 N
Answer:
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Explanation:
