A basketball rolls without slipping (starting from rest) down a ramp. If the ramp is sloped by an angle of 4 degrees above the h
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Answer:
maximum speed of the block is 3.14 m/s
Explanation:
given data
mass = 1.70 kg
stretch in the spring = 0.2 m
time take by block to come to zero t = 0.2 s
solution
we know that Time period of oscillation (T) that is express as
T = 2t ......................1
put here value
T = 2 (0.2)
T = 0.4 s
so here time period is express as
T = ................2
here k is spring constant of the spring so put here value
0.4 =
here k will be
k = 419.02 N/m
so we use here conservation of energy that is
Maximum kinetic energy = Maximum spring potential energy ............3
(0.5) m v² = (0.5) k x²
here v is maximum speed block
so put here value and we get
(1.70) v² = (419.02) (0.2)²
v = 3.14 m/s
so maximum speed of the block is 3.14 m/s
Answer:
100 m/s
Explanation:
Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁
Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is completely stationary. Let v₁ is speed of bond's boat.
It is the concept of the conservation of momentum. It remains conserved. So,
Putting all the values, we get :
So, Bond's boat is moving with a speed of 100 m/s after the collision.