A. The number of files was constant.
Answer:
Explanation:
When he accelerates the heavy object up , the reading increases because an extra downward normal force acts on it, then scale reading returns to the same reading as when standing stationary, and then decreases as although he is lifting the heavy object , the acceleration is decreasing ,so the extra upward normal force acts.
Answer:
3.7 m
Explanation:
ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky
The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of
t = √ (2h/g) = √(2(15)/9.8) = 1.75 s
Without the tail wind, the ball travels horizontally
d = vt = 68(1.75) = 119 m
The tailwind exerts a constant acceleration on the ball of
a = F/m = 12/5.0 = 2.4 m/s²
The average horizontal velocity during the flight is
v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s
so the distance with tailwind is
d = v(avg)t = 70.1(1.75) = 122.675 m
The extra distance is 122.675 - 119 = 3.675 = 3.7 m
Where speed is distance/time, velocity is displacement/time.
What this means is that velocity is the length covered in relation to the starting point.
Speed is just the distance travelled no matter where you began.
When going around a circular track, you might have a speed value. However, since you get back to the same location at every lap, you have 0 velocity.
Hope I helped :)
Answer:
Explanation:
Given,
- Work done by the rope 900 m/s.
- Angle of inclination of the slope =
- Initial speed of the skier = v = 1.0 m/s
- Length of the inclined surface = d = 8.0 m
part (a)
The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity
In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.
part (b)
- Initial speed of the skier = v = 1.0 m/s.
Rate of the work done by the rope is power of the rope.
Part (c)
- Initial speed of the skier = v = 2.0 m/s.
Rate of the work done by the rope is power of the rope.
