Do you remember this formula for the distance traveled while accelerated ?
<u>Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)</u>
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice <em>D)</em> is the closest one.
<span>When an atomic nucleus emits a beta particle, "Atomic number remains same"
Hope this helps!</span>
Answer:
10.77m
Explanation:
The elastic potential energy of the spring when compressed with the mass is converted to the kinetic energy of the mass when released. This ie expressed in the following equation;

where k is the force constant of the spring, e is the compressed length, m is the mass of the block and u is the velocity with which the block leaves the spring after being released.
If we make u the subject of formula from equation (1) we obtain the following;

Given;
e = 0.105m,
k = 4825N/m,
m = 0.252kg,
u = ?
Substituting all values into equation (2) we obtain the following;

The maximum height attained is then obtained from the third equation of motion as follows, taking g as 

v = 0m/s
Hence

Answer:0.375m3
Explanation:This is general gas law. The expression is given by P1V1/T1 =P2V2/T2
we are looking for V2 so let's make it the change of subject of formula
V2= P1xV1xT2/P2T1
= 5 x 0.6 x 200/400 x 4
= 0.375m3