Answer:
8.5 m/s
Explanation:
please see paper for the work!
Answer:

Explanation:
When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

Where:
vf = final velocity
vo = initial velocity
a = acceleration
x = distance
Solving for a:

The car travels a distance of x=200 m and the velocities are:
vo = 216 Km/h
vf = 360 Km/h
Both velocities must be converted to meters by seconds.
vo = 216 Km/h *1000/3600 = 60 m/s
vf = 360 Km/h *1000/3600 = 100 m/s
The acceleration is:




Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
Answer:
Explanation:
You don't do the same amount of work. The work formula is F*d = W
W = work
F = force
d = the distance moved.
So you do more work when you move the 10N object because the Force (F) has doubled.
Decibels I believe? I’m not 100% sure