Answer:
Mass of bullet is m=0.01kg
Mass of the block is M=4kg
Coefficient=0.25,distance=20m
So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,
By applying conservation of momentum,
mv=(m+M)V
V=
M+m
mv
Explanation:
please mark me as the brainliest answer and please follow me for more answers to your questions..
If there was any way to do that, then your teacher wouldn't
need to keep you coming into class every day and doing
homework every night. She could just give you the 3 or 4
paragraphs and a few pictures that you're asking me for,
and bada-bing ! you'd know it !
The time it takes, and the amount of homework it takes, is
EXACTLY the time you spent hearing about it in class.
(Unless you're some kind of genius savant prodigy, which
you're not and I'm not.)
Answer:
By preventing the government from taking property without fair payment.
Explanation:
Answer 1
Please mark me as brilliant and thank you
The last choice. Two arrows and the arrow up is shorter than the arrow down. Since the guy is falling and he’s opened his chute, he’s slowing down but he’s still falling meaning the force of gravity is stronger than the air resistance.
Answer:
v = √[gR (sin θ - μcos θ)]
Explanation:
The free body diagram for the car is presented in the attached image to this answer.
The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.
Resolving the weight into the axis parallel and perpendicular to the inclined plane,
N = mg cos θ
And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ
Frictional force = Fr = μN = μmg cos θ
Centripetal force responsible for keeping the car in circular motion = (mv²/R)
So, a force balance in the plane parallel to the inclined plane shows that
Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)
(mv²/R) = (mg sin θ - μmg cos θ)
v² = R(g sin θ - μg cos θ)
v² = gR (sin θ - μcos θ)
v = √[gR (sin θ - μcos θ)]
Hope this Helps!!!