Answer:
630.75 j
Explanation:
from the question we have the following
total mass (m) = 54.5 kg
initial speed (Vi) = 1.4 m/s
final speed (Vf) = 6.6 m/s
frictional force (FF) = 41 N
height of slope (h) = 2.1 m
length of slope (d) = 12.4 m
acceleration due to gravity (g) = 9.8 m/s^2
work done (wd) = ?
- we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy
wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)
wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)
where wd = work done
m = mass
h = height
g = acceleration due to gravity
FF = frictional force
d = distance
Vf and Vi = final and initial velocity
wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)
wd = 630.75 j
The answer for this is b 3.500.000j
Answer:Decreases
Explanation:
Given
Volume is held constant that is it is a isochoric process.
We know that
PV=nRT
as n,V& R are constant therefore only variables are
P & T
so 

As
is decreasing therefore Pressure must also decrease so that ratio remains constant.
Speed with which initially car is moving is 21 m/s
Reaction time = 0.50 s
distance traveled in the reaction time d = v t
d = 21 * 0.50 = 10.5 m
deceleration after this time = -10 m/s^2
now the distance traveled by the car after applying bakes



so total distance moved before it stop
d = 22.05 + 10.5 = 32.55 m
so the distance from deer is 35 - 32.55 = 2.45 m
now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop
so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m
again by kinematics



so maximum speed would be 22.1 m/s