<h2>
Height of cliff is 66.43 meter.</h2>
Explanation:
Consider the horizontal motion of shell,
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 1770 m/s
Acceleration, a = 0 m/s²
Displacement, s = 6520 m
Substituting
s = ut + 0.5 at²
6520 = 1770 x t + 0.5 x 0 xt²
t = 3.68 s
Time of flight of shell = 3.68 s.
Now consider the vertical motion of shell,
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3.68 s
Substituting
s = ut + 0.5 at²
s = 0 x 3.68 + 0.5 x 9.81 x 3.68²
s = 66.43 m
Height of cliff is 66.43 meter.
The answer is: The addition of a small dense nucleus at the center of the atom.
Answer:
(a). The required is 1871.2 J.
(b). The speed the lead bullet is 395 m/s.
(c). The 20% of energy must have gone into collision between steel plate and bullet.
Explanation:
Given that,
Mass of bullet = 0.030 kg
Temperature = 20°C
(a). We need to heat required to increase the temperature of the lead bullet and melt it
Using formula of heat
Where, m = mass of lead bullet
S = specific heat
L = latent heat
T = Temperature
Put the value into the formula
The required is 1871.2 J.
(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it
We need to calculate the energy
We need to calculate the speed the lead bullet
Using formula of speed
The speed the lead bullet is 395 m/s.
(c). The 20% of energy must have gone into collision between steel plate and bullet.
Hence, This is the required solution.
Answer:
Acceleration due to gravity is 20 
So option (E) will be correct answer
Explanation:
We have given length of the pendulum l = 2 m
Time period of the pendulum T = 2 sec
We have to find acceleration due to gravity g
We know that time period of pendulum is given by
Squaring both side
So acceleration due to gravity is 20
So option (E) will be correct answer.
Answer:
b friction
Explanation:
Contact forces
Contact forces are forces that act between two objects that are physically touching each other. Examples of contact forces include:
Reaction force
An object at rest on a surface experiences reaction force. For example, a book on a table.
A box rests on a table. There are two arrows, equal in size but going in opposite directions, up and down, from the point where the box meets the table.
Tension
An object that is being stretched experiences a tension force. For example, a cable holding a ceiling lamp.
A box hangs from a rope. Two arrows which are equal in size act upwards and dowards from the top and bottom of the rope.
Friction
Two objects sliding past each other experience friction forces. For example, a box sliding down a slope.
A box rests on an incline. There are three arrows; one acting vertically downwards from the centre of the box’s base. One arrow acts perpendicular to the incline. One arrow acts up the incline.
Air resistance
An object moving through the air experiences air resistance. For example, a skydiver falling through the air.
A box falls from the sky. Two arrows, equal in size and opposite in direction act upwards from the box and downwards from the box
When a contact force acts between two objects, both objects experience the same size force, but in opposite directions. This is Newton's Third Law of Motion.
