Explanation:
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Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
v = 2 m/s
Hence, their speed after collision is 2 m/s.
Answer:
Sum of the forces will be equal to 3.479 N
Explanation:
We have given two same forces are oriented at an angle of 38°
Magnitude of each force is given
We have to find the sum of the forces
Sum of the forces will be equal to
So sum of the forces will be equal to
So sum of the force will be equal to 3.479 N
v = initial velocity of launch of the stone = 12 m/s
θ = angle of the velocity from the horizontal = 30
Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.
v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s
a = acceleration of the stone = - 9.8 m/s²
t = time of travel = 4.8 s
Y = vertical displacement of stone = vertical height of the cliff = ?
using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²
Y = - 84.1 m
hence the height of the cliff comes out to be 84.1 m
Answer:
20m/s
Explanation:
acceleration=final velocity-initial velocity/time
4.0m/s²=v m/s-0m/s/5.0sec
5.0sec×4.0m/s²=v m/s-0m/s×5.0m/s/5.0m/s
20m/s=v