<span>95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M</span>
work done is product of force and displacement of point of application of force
so here we have to check the product of force and displacement both
Now we will put the least to maximum work in the following order
1. -A man exerts strenuous effort in pushing a stationary wall
2. -A flea pushes a speck of dirt 1 cm
3. -A farmer pushes a 2 kg wheelbarrow 20 m
4. -A cannon launches a 3 kg cannonball a distance of 200
5. -A 2000 kg car travels 400 m down a road
6. -Space shuttle Atlantis launches from the ground into near-Earth orbit
Hello!!
Here we have a simple matter of conservation of energy. ME=PE+KE.
At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.
Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!
Answer:
Part a)
Part B)
Explanation:
As we know that when both the forces are acting on the object in same direction then we will have
as we know that
m = 10.6 kg
now we will have
Now two forces are in opposite direction then we have
Part A)
Now we will have from above two equation
Part B)
Similarly for other force we have
Is there supposed to be an image if so there is none
