Answer:
d = 506.25 ft
Explanation:
As we know by kinematics that
here we know that initially the stone is dropped from rest from the edge of the roof
so here initial speed will be zero
now we have
also the acceleration of the stone is due to gravity which is given as
now we have
so from above equation
Answer: 33 mm
Explanation:
Given
Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m
Internal pressure of gas, P(i) = 1.5 MPa
Yield strength of steel, P(y) = 340 MPa
Factor of safety = 0.3
Allowable stress = 340 * 0.3 = 102 MPa
σ = pr / 2t, where
σ = allowable stress
p = internal pressure
r = radius of the tank
t = minimum wall thickness
t = pr / 2σ
t = 1.5*10^6 * 4.5 / 2 * 102*10^6
t = 0.033 m
t = 33 mm
The minimum thickness of the wall required is therefore, 33 mm
The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.
This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.
The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.
Answer:
Force on the object is 20 N
Explanation:
As we know that work done to raise the book from initial position to final position is known as potential energy stored in it
So here we know that
here we know that
U = 30 J
s = displacement = 1.5 m
so we have
Answer:
The value of change in internal energy of the gas = + 1850 J
Explanation:
Work done on the gas (W) = - 1850 J
Negative sign is due to work done on the system.
From the first law we know that Q = Δ U + W ------------- (1)
Where Q = Heat transfer to the gas
Δ U = Change in internal energy of the gas
W = work done on the gas
Since it is adiabatic compression of the gas so heat transfer to the gas is zero.
⇒ Q = 0
So from equation (1)
⇒ Δ U = - W ----------------- (2)
⇒ W = - 1850 J (Given)
⇒ Δ U = - (- 1850)
⇒ Δ U = + 1850 J
This is the value of change in internal energy of the gas.
