Answer:
12.65 m/s
Explanation:
When friction is ignored, if the car travels from 18m hill to the bottom of the hill, afterwards it travels up to a 10m hill, then overall it has traveled a distance of 18 - 10 = 8 m vertically.
If the car starts from rest, then its potential energy is converted to kinetic energy:
where m is the car mass and h is the vertical distance traveled, v is the car velocity at 10m hill
Let g = 10m/s^2. We can cancel m from both sides of the equation:
Answer:
640 m.
Explanation:
The following data were obtained from the question:
Acceleration (a) = –20 m/s/s
Time (t) = 8 s
Final velocity (v) = 0 m/s
Distance (s) =.?
Next, we shall determine the initial velocity (u) of the car. This can be obtained as follow:
Acceleration (a) = –20 m/s/s
Time (t) = 8 s
Final velocity (v) = 0 m/s
Initial velocity (u)
a = (v – u) / t
–20 = (0 – u) / 8
–20 = – u / 8
Cross multiply
–20 × 8 = – u
– 160 = – u
Divide both side by – 1
u = – 160 / – 1
u = 160 m/s
Finally, we shall determine the distance travelled by the car before stopping as follow:
Time (t) = 8 s
Final velocity (v) = 0 m/s
Initial velocity (u) = 160 m/s
Distance (s) =.?
s = (v + u)t /2
s = (0 + 160) × 8 /2
s = (160 × 8) /2
s = 1280 / 2
s = 640 m
Therefore, the car travelled 640 m before stopping.
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
53
i did it and got it right guys i hope it helps a lot
Answer:
Explanation:
Let mass of ice required is m kg
Heat gained by ice to attain zero degree
= m x 14.1 x 2100
= 29610m J
Heat gained by ice to melt = m x 3.34 x 10⁵ J
= 334000m J
Heat gained by water at zero degree to warm up to 27 degree
m x 4190 x 27 =113130 J
Total heat gained = 476740m J
Heat lost by hot water to cool up to 27 degree
.24 x 4190 x ( 75.8 - 27 )
= 49073.28 J
Heat lost = heat gained
476740 m = 49073.28
m = 49073.28/476740 kg
=102.93 gm