Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
Answer:
Earth's interior (Core)
Explanation:
The earth is comprised of 3 distinct layers namely the Core, the Mantle and the Crust, which are divided based on their composition as well as density.
The core of the earth is extremely very hot where the inner core remains solid and outer core acts a liquid. It is mainly comprised of iron, nickel and other siderophile elements.
A large amount of heat (energy) is radiated from this core region towards the surface of the earth. Due to this, the mantle rocks forms magma that creates the convection currents, where the hot and less dense magma rises upward and the cool and denser magma sinks to the bottom. This occurs continuously, as a result of which the lithospheric plates are forced to move over the less dense layer of asthenosphere.
Thus, the heat energy that drives the convection current in the mantle is provided from the interior (core) of the earth.
Answer:
3) D: 31 m/s
4) D: 84.84 metres
Explanation:
3) Initial velocity along the x-axis is;
v_x = v_o•cos θ
Initial velocity along the y-axis is;
v_y = v_o•sin θ
Plugging in the relevant values, we have;
v_x = 31 cos 60
v_x = 31 × 0.5
v_x = 15.5 m/s
Similarly,
v_y = 31 sin 60
v_y = 31 × 0.8660
v_y = 26.85 m/s
Thus, magnitude of the initial velocity is;
v = √(15.5² + 26.85²)
v ≈ 31 m/s
4) Formula for horizontal range is;
R = (v² sin 2θ)/g
R = (31² × sin (2 × 60))/9.81
R = 84.84 m
Answer: Option (b) is the correct answer.
Explanation:
It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.
Then test charge to the right immediately after being released.
Therefore, the net force will be as follows.
F = 
= 
= 
F = > 0
Thus, we can conclude that the test charge move to the right immediately after being released.
Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
