Answer:
a = 3.27 m/s²
v = 2.56 m/s
Explanation:
given,
mass A = 1 kg
mass B = 2 kg
vertical distance between them = 1 m
a = 3.27 m/s²
The speed of the system at that moment is:
v² = u² + 2×a×s
v² = 0² + 2× 3.27 × 1
v ² = 6.54
v = 2.56 m/s
The answer is : C ) air,water,and the tank glass.
-Hope this helps.
Answer:
U₂ = 400 KJ
Explanation:
Given that
Initial energy of the tank ,U₁= 800 KJ
Heat loses by fluid ,Q= - 500 KJ
Work done on the fluid ,W= - 100 KJ
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take final internal energy =U₂
We know that
Q= U₂ - U₁ + W
-500 = U₂ - 800 - 100
U₂ = -500 +900 KJ
U₂ = 400 KJ
Therefore the final internal energy = 400 KJ
The two types of motion exerted in bicycle are:
1. rotary motion
2. linear motion
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second