(a) The momentum of the first trolley is 5.4 kgm/s
(b) The velocity of the trolleys after impact is 2.7m/s
<u>Explanation:</u>
Given:
Mass, m₁ = 1.2kg
Velocity, v₁ = 4.5m/s
Mass, m₂ = 0.8kg
v₂ = 0
(a) Momentum of the trolley before impact, p
We know:
Momentum = mass X velocity
p = 1.2 X 4.5
p = 5.4 kgm/s
Therefore, the momentum of the first trolley is 5.4 kgm/s
(b) Speed of the trolleys after impact, v = ?
During collision, the momentum is conserved.
So,
m₁v₁ + m₂v₂ = (m₁ + m₂)v
(1.2 X 4.5) + (0.8 X 0) = (1.2 +0.8) X v
5.4 + 0 = 2v
v = 2.7m/s
Therefore, the velocity of the trolleys after impact is 2.7m/s
Answer:
E ) The horizontal component of a projectile acceleration is zero.
Explanation:
In case of a projectile , force of gravity acts in vertically downward direction so acceleration will act in vertically downward direction . Its direction never changes during course of its journey. So horizontal component of acceleration will always be zero at all points of its journey.
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100
degrees Celsius.
Required:
Heat produced
Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules
per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat
produced in boiling 2 kg of water.
Answer:
196
Explanation:
subtract 24 from 220 to get your answer.
