1. A force of 25.0 Newtons is applied so as to move a 5.0 kg mass a distance of 20.0 meters. How much work was done?
Ans. W = F × d = 25 N × 20.0 m = 50 J
2. A force of 120 N is applied to the front of a sled at an angle of 28.0 above the horizontal so as to pull the sled a distance of 165 meters. How much work was done by the applied force?
Ans. W = F •d cos = 120 N • 165 m cos 28.0 = 17,482.36 J
3. A sled, which has a mass of 45.0 kg., is sitting on a horizontal surface. A force of 120 N is applied to a rope attached to the front of the sled such that the angle between the front of the sled and the horizontal is 35.0o. As a result of the application of this force the sled is pulled a distance of 500 meters at a relatively constant speed. How much work was done to this sled by the applied force?
Ans. W = F• d cos = 120 N • 500 m • cos 35.0 = 49,149.12 N
4. A rubber stopper, which has a mass of 38.0 grams, is being swung in a horizontal circle which has a radius of R = 1.35 meters. The rubber stopper is measured to complete 10 revolutions in 8.25 seconds.
a. What is the speed of the rubber stopper?
Ans. v = v = • v = .•(. ) v = 10.29 m/s .
b. How much force must be applied to the string in order to keep this stopper moving in this circular path at a constant speed?
Ans. The Centripetal Force, Fc = • = . • (. /)= 29.76 N .
c. How far will the stopper move during a period of 25.0 seconds?
Ans. d = vt = 10.29 m/s25 s = 257.25 m
d. How much work is done on the stopper by the force applied by the string during 25.0 seconds?
Ans. 0 J. The force is towards the center and the distance is around the perimeter of the circle. They are at right angles to each other. For work to be done, the force has to be in the same direction as the displacement.
5. How much work would be required to lift a 12.0 kg mass up onto a table 1.15 meters high?
Ans. W = F• d = mg d = 12 kg 9.8 m/s2 1.15 m = 135.24 J
Answer:
one half as large , initial velocity is two times larger
Explanation:
Momentum is conserved.
p₁ + p₂ = p₁' + p₂'
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
m₁ = m₂ =m , v₂= 0
v₁' =v₂'
mv₁ = 2mv₁'
v₁ = 2v₁'
D)<span>As the internal energy increases a substance would go from solid to a liquid.</span>
Answer:
image is vertical at distance -203.62 cm
magnification is 2.110
Explanation:
given data
n = 1.51
distance u = 96.5 cm
concave radius r1 = 24 cm
convex radius r2 = 19.1 cm
to find out
final image distance and magnification
solution
we will apply here lens formula to find focal length f
1/f = n-1 ( 1/r1 - 1/r2) .......................1
put here all value
1/f = 1.51 -1 ( -1/24 + 1/19.1)
f = 183.43
so from lens formula
1/f = 1/v + 1/u .............................2
put here all value and find v
1/183.43 = 1/v + 1/96.5
so
v = −203.62 cm
so here image is vertical at distance -203.62 cm
and
magnification are = -v /u
magnification = 203.62 / 96.5
magnification is 2.110
