Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories. Not only is measurement important in science and the chemical industry, it is also essential in farming, engineering, construction, manufacturing, commerce, and numerous other occupations and activities.
use F = ma
F : force m : mass a : acceleration
so
f = 5kg * 20 m/s2 = 100 N
Answer:
1. P⃗ A,i+P⃗ B,i=P⃗ A,f+P⃗ B,f
Explanation:
Collision occurs when two bodies moving at different velocities exert force on each other through colliding. Collision can either be elastic or inelastic.
For elastic collision, both energy and momentum of the bodies is conserved i.e they separates after collision.
For inelastic collision, only momentum is conserved but not energy. The body sticks together after colliding and moves with a common velocity.
According law of conservation of momentum which states that the sum or momentum of two bodies before collision is equal to the momentum of the bodies after collision.
Given two bumper cars A and B that undergoes collision during which momentum is conserved, the type of collision that exists between the bumpers is inelastic collision
.
If initial momentum of A is PiA and initial momentum of B is PiB, the sum of their momentum before collision is expressed as;
PiA + PiB ... (1)
If final momentum of A is PfA and initial momentum of B is PfB, the sum of their momentum after collision is expressed as;
PfA+PfB ... (2)
According to the law, equation 1 is equal to equation 2
The fore the formula that states the law is expressed as;
PiA + PiB = PfA+PfB
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
