Answer:
maybe the rear wheel axel rusted not sure
By definition of average acceleration,
<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²
Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:
∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N
If you just want the magnitude, drop the negative sign.
The things that determine the amount of an object's gravitational potential energy are ...
-- mass of the object
-- gravitational acceleration in the place where the object is
-- height of the object above the ground or the floor
A). <em>a slice of bread; </em> No. It's still a slice of bread even if it's on the ground.
B. <em>A compressed spring; </em> No. It's still a compressed spring even if it's on the ground.
C. <em>An apple on a tree</em>; <em>Yes !</em> It can't be an apple on a tree if it's on the ground.
D. <em>A stretched bow string</em>; <em>No.</em> It's still a stretched bowstring even if it's on the ground.
Answer:
equivalent power = 6.60 × 
W
Explanation:
given data
coal train length = 1.5 km = 1500 m
contain = 120 cars
holding= 110 tonnes of coal each
train traveling speed = 100 km/h = 27.78 m/s
to find out
equivalent power in watts
solution
we take here energy produce by coal is 27 × 
J/kg
and here total coal = 120 × 110 × 10³ kg
sop here energy produce by total coal is
energy produce by total coal = 27 × × 120 × 110 × 10³
and time required to cross that distance is
time = 
time = 
time = 53.99 sec
so here equivalent power will be
equivalent power = 
equivalent power = 
equivalent power = 6.60 × W
Answer:
9.5 m/s
Explanation:
Distance, S = 150m
Acceleration, a = 0.3 m/s^2
Initial velocity, u = 0 m/s
Final velocity, v
Use kinematics equation
v^2 - u^2 = 2aS
v^2 - 0 = 2*0.3*150 = 90
v = sqrt(90) = 9.49 m/s
