Answer:
Explanation:
As we know that at isochoric conditions if pressure of the gas is tripled then the temperature also becomes 3 times
So heat given in this process is given as
Now it is expanded to initial pressure again by adiabatic process
So in this part there is no heat exchange
Also we know that
Now in the last process we compressed the gas to original volume
Now total heat in the process is given as
Work done = 1/2*(max. force - min. force) * greatest extenstion
Max. force = spring constant * greatest extension= 80*d = 80d N
Min. force = spring constant * smallest extenstion = 80*0 = 0 N
Therefore,
Work done = 1/2*80d*d = 40d^2 J
However,
Mechanical energy = Work done
That is,
0.12 = 40d^2
d = Sqrt (0.12/40) = 0.0548 m
The greatest extension from its equilibrium is 0.0548 m.
Answer:
The Tension T is 42120N
The Horizontal force component is 18322.2N
The Vertical force component is - 4729N
Explanation:
First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).
Having done that, you apply two conditions of equilibrium.
1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.
∑Fx = 0
∑Fx = Rx - Tcos64.2 = 0
Rx = 0.435T
∑Fy = 0
∑Fy = Ry + Tsin64.2 - W - w = 0
W = 2000kg × 9.8 = 19600N
w =1000kg × 9.8 = 9800N
Ry + 0.9T = 29400N
Ry = 29400 - 0.9T
2. THE SUM TOTAL OF TORQUES EQUALS ZERO
Rx: τ = 0
Ry: τ = 0
T: τ = 5 × Tsin44.2
= 3.49T m
W: τ = 4 × 19600sin90
= 78400Nm
w: τ = 7 × 9800sin9
= 68600Nm
Note:
Rx = x component of Reaction force
Ry = y component of Reaction force.
T = Tension
W = weight of bridge
w = weight of Sir Lance a Lost and his steed
τ = torque
Note: The torque of Tension is counter clockwise while that of the weights is clockwise.
Hence,
∑τccw = ∑τcw
3.49T = 78400 + 68600
3.49T = 14700Nm
T = 147000/3.49
T = 42120N
Rx = 0.435 × 42120
Rx = 18322.2N
Ry = 29400N - (0.9×42120)N
Ry = 29400 - 34129
Ry = -4729N
Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.
Answer:
<em>The rebound speed of the mass 2m is v/2</em>
Explanation:
I will designate the two masses as body A and body B.
mass of body A = m
mass of body B = 2m
velocity of body A = v
velocity of body B = -v since they both move in opposite direction
final speed of mass A = 2v
final speed of body B = ?
The equation of conservation of momentum for this system is
mv - 2mv = -2mv + x
where x is the final momentum of the mass B
x = mv - 2mv + 2mv
x = mv
to get the speed, we divide the momentum by the mass of mass B
x/2m = v = mv/2m
speed of mass B = <em>v/2</em>