Answer:
There is no great force, the force exerted by the Earth on the Sun, and the force exerted by the Sun on the Earth are equal
Explanation:
By definition...
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F = 
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
Yes it does, uh huh. It slows down as it rolls. That's a fact.
In order for the ball to roll forward, it has to push grass out of the way. That takes energy. To bend each blade of grass out of its way, the ball has to use a tiny bit of the kinetic energy that it has, so it gradually runs out of kinetic energy. When its kinetic energy is all gone, it stops moving.
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:
brainly.com/question/25573309
#SPJ1
Answer:
2156 J
Explanation:
From the question,
Work done = Combined mass of the bucket and water×height×gravity.
W = (M+m)hg............................. Equation 1
Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.
Given: M = 20 kg, m = 2 kg, h = 10 m
Constant: g = 9.8 m/s²
Substitute these value into equation 1
W = (20+2)×10×9.8
W = 22×98
W = 2156 J
