The total work done on the car is 784Joule.
<h3>What's the acceleration of the car?</h3>
- As per Newton's equation of motion, V= U+at
- U= initial velocity= 0 m/s
V= vinal velocity= 20m/s
t= time = 10s
a= acceleration
=> a= 20/10= 2m/s²
<h3>What's the distance covered by the car in 10 seconds?</h3>
- As per Newton's equation of motion,
V²-U² = 2aS
- S= distance covered by the car
- So, 20²-0=2×2×S=4S
=> 400= 4S
=> S= 400/4= 100m
<h3>What's the work done on the car due to frictional force?</h3>
Work done by frictional force= frictional force × distance
= (0.2×4×9.8)×100
= 784Joule
Thus, we can conclude that the work done on the car is 784Joule.
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Answer:
constant at the speed of light
Answer:
The force required to move the quarterback with linebacker is <u>1215 N</u>
Explanation:



Using Newton's second law, it is established that F = Ma
Where F is net force acting on the system, a is the acceleration and M is mass of the two object 
Now consider both
as a system, so net force acting on the system is 
Substitute the given values in the above formula,


Force = 1215 N
<u>1215 N </u>is the force required to move the quarterback with linebacker.
Use the eq. of Young modulus Y=(F/A)/(∆l/lo)
dimana ∆l is the elongation of wire, lo is its initial length.
So ∆l = (F/A)lo/Y.
∆l = (1000N/(6.5 × 10^-7 m^2))×(2.5m)/(2.0 × 10^-11 N/m^2)
Use calculator to finish it.