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.
Learn more about the work done here:
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W = 4.9N. The weight of a basketball with a mass of 0.5Kg is 4.9N.
The weight of an object is the force of gravity on the object and can be defined as the product of the mass by the acceleration of gravity, w = mg.
W = (0.5Kg)(9.8 m/s²) = 4.9N
Answer:
The car would speed off 2x's as fast as the speed of the heavy truck provided the the collision is an elastic collision where there's no or little friction occurring within the scenario.
Explanation:
Newton's law proves that an object with a greater mass can move objects of lesser mass at greater distances and speed.
Answer:
1.147 %
Explanation:
= Density of mercury = 
g = Acceleration due to gravity = 9.81 m/s²
h = Height of mercury = 8.69 m
Standard atmospheric pressure = 
6.55 mmHg converting to Pa
Dividing air pressure by the above value
The fraction of atmospheric pressure is 1.147 %
