Balance and beaker of water. The balance will measure mass and the beaker will measure the volume when you take the initial volume without the key subtracting that value from the value with the key in the beaker
Answer:
4948020
Explanation:
(6.9*10-6)(770*102)
<u>Multiply 6.9 by 10
.
</u>
(
69
−
6
)
(
770
⋅
102
)
<u>Subtract 6 from 69
.
</u>
63
(
770
⋅
102
)
<u>Multiply 770 by 102
.
</u>
63
⋅
78540
<u>Multiply 63 by 78540
.
</u>
4948020
Answer:
a). The velocity of the first log is -1.65 m/s.
(b). The velocity of the second log is 1.07 m/s.
Explanation:
Given that,
Mass of lumberjack M= 110 kg
Mass of log m= 206 kg
Final velocity = 3.09 m/s
(a). We need to calculate the velocity of the first log just before the lumberjack jumps off
Using conservation of momentum
Put the value into the formula
The velocity of the first log is -1.65 m/s.
(b). If the lumberjack comes to rest relative to the second log
We need to calculate the velocity of the second log
Put the value into the formula
The velocity of the second log is 1.07 m/s.
Hence, (a). The velocity of the first log is -1.65 m/s.
(b). The velocity of the second log is 1.07 m/s.
Our solar system is the only place in the universe where gravity played a key part in the formation of planets.
This is Flase
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
