C
Unbalanced causes something to move because the net force is greater than zero
Answer:
A fair test.
Explanation:
Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.
In a fair test two or more things are compared.
It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.
The factor is called a variable.
Answer: 247.67 V
Explanation:
Given
Potential At A 
Potential at 
when particle starts from A it reaches with velocity
at Point while when it starts from C it reaches at point B with velocity 
Suppose m is the mass of Particle
Change in Kinetic Energy of particle moving under the Potential From A to B

Change in Kinetic Energy of particle moving under the Potential From C to B

Divide 1 and 2 we get

on solving we get


Net force on the car=F=4.8 x 10³ N
Explanation:
mass of car= 1.2 x 10³ Kg
initial velocity= Vi=0
Final velocity= Vf= 20 m/s
time = t= 5 s
Using kinematic equation,
Vf= Vi + at
20= 0 + a (5)
5 a=20
a= 20/5
a= 4 m/s²
Now force is given by F = ma
F= 1.2 x 10³ (4)
F=4.8 x 10³ N
Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .