Answer is (c), the latent heat of fusion. That is by definition the heat that 1 kg of a substance must absorb to melt in the vicinity of its melting point.
Answer:
The false statement is in option 'd': The center of mass of an object must lie within the object.
Explanation:
Center of mass is a theoretical point in a system of particles where the whole mass of the system is assumed to be concentrated.
Mathematically the position vector of center of mass is defined as

where,
is the position vector of the mass dm.
As we can see for homogenous symmetrical objects such as a sphere,cube,disc the center of mass is located at the centroid of the shapes itself but in many shapes it is located outside the body also.
Examples of shapes in which center of mass is located outside the body:
1) Horseshoe shaped body.
2) A thin ring.
In many cases we can make shapes of bodies whose center of mass lies outside the body.
Answer:
B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.
Explanation:
In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential
so according to the given situation, the option b is correct
And, the rest of the options are wrong
Answer:

Explanation:
Using the conservation of energy we have:

Let's solve it for v:

So the speed at the lowest point is 
Now, using the conservation of momentum we have:

Therefore the speed of the block after the collision is 
I hope it helps you!
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft