Answer:
v₂ = 22.5 m/s
Explanation:
Given that
For puck 1
m₁= 1 kg
u₁= 30 m/s
For puck 2
m₂= 1 kg
u₂= 0 m/s
After collision
Puck 1 have velocity v₁=7.5 m/s
Take puck 2 will have velocity v₂
From linear momentum conservation
P₁=P₂
m₁ u₁+m₂ u₂=m₁ v₁+m₂ v₂
1 x 30 + 1 x 0 = 1 x 7.5 + 1 x v₂
30 - 7.5 =v₂
v₂ = 22.5 m/s
20h
Explanation:
Given parameters:
Electrical power = 75W
Electrical energy = 1500J
Unknown:
Time taken = ?
Solution:
Electrical power is the amount of voltage that drives current.
P = IV
Electrical energy is the voltage that drives a current per unit of time
E = IVt
E = Pt
Since time is the unknown, we can make it subject of the formula;
t =
=
= 20h
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Invention of electrical power brainly.com/question/13146914
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Answer:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ

At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
A series circuit is a closed circuit in whicj the current flows through one path.