A 17.5-kg block is dragged over a rough, horizontal surface by a 75-N force acting at 21° above the horizontal. The block is dis
placed 5.7 m, and the coefficient of kinetic friction is 0.150.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.
1 answer:
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Answer:
27.95[kW*min]
Explanation:
We must remember that the power can be determined by the product of the current by the voltage.
where:
P = power [W]
V = voltage [volt]
I = amperage [Amp]
Now replacing:
Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]
Answer:
4v/3
Explanation:
Assume elastic collision by the law of momentum conservation:
where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively
Substitute
Divide both side by , then multiply by 6 we have
So the final speed of the second car is 4/3 of the first car original speed