For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2
If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2
Answer: The scientist gives up and starts an investigation on a new topic.
Explanation:
The data is altered so that it supports the original hypothesized. The data is then altered so that it supports the original hypothesis.
Answer:
Final angular velocity is 35rpm
Explanation:
Angular velocity is given by the equation:
I1w1i + I2w2i = I1w1f -I2w2f
But the two disks are identical, so Ii =I2
wf can be calculated using
wf = w1i - w2i/2
Given: w1i =50rpm w2i= 30rpm
wf= (50 + 20) / 2
wf= 70/2 = 35rpm
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11