Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
Power used by the clock=1.03 W
Explanation:
resistance= 14000 ohm
voltage=120 V
The formula for the power is given by

P=(120)²/14000
P=1.03 W
93.5 if it’s wrong sorry sis I need my homework done too
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/
, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut +
a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s =
a
=
x 14 x 
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.