Answer:
-223.64684 J
Explanation:
F = Force that is applied to the crate = 68 N
s = Displacement of the crate = 3.5 m
= Angle between the force and displacement vector = (180-20)
Work done is given by

The work that Paige does on the crate is -223.64684 J
Answer:
b. 46 km/hr
Explanation:
Applying,
S' = d'/t'................... Equation 1
Where S' = Average speed of the bus, d' = Total distance covered by the bus, t' = Total time taken.
From the question,
d' = (100+55+75) km = 230 km
t' = (2+1+2) = 5 hours
Substitute these values into equation 1
S' = 230/5
S' = 46 km/hr.
Hence the correct option is b. 46 km/hr
Answer:
Explanation:
The sig figs are off in the answer (unless you use -9.81 for the acceleration due to gravity. My classes always use -9.8, which is 2 sig figs. But nevertheless, we can solve it!) Use the velocity formula:
where vf is the final velocity, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time in seconds. Filling in, using the fact that both the velocities are negative since the ball is going down:
-24.5 = -12.1 +(-9.8)t or
-24.5 = -12.1 - 9.8t and
-12.4 = -9.8t and divide to get that
t = 1.27 sec
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s