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Paraphin [41]
3 years ago
7

In a lab,two pieces of cellphane tape are placed on a lab table. They are quickly ripped off the lab table. Which would you pred

ict would happen if the two pieces of tape were brought near each other?
Physics
1 answer:
galben [10]3 years ago
4 0

Answer:

they would repel each other

Explanation:

they have gained the same charge

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Would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that was
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More compressed. moving up = apparent weight (i.e., your norma force) is greater. this means you’ll weighr more and push those springs down even more than you would at rest.
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3 years ago
A worker pushes horizontally on a large crate with a force of 200 N, and the crate is moved 3.5 m. How much work was done in Jou
sattari [20]

Work = (force) x (distance) =

           (200 N) x (3.5 m)  =  <em>700 joules</em>


6 0
4 years ago
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5 0
3 years ago
Is it ever possible for the work done by friction to increase the kinetic energy of an object?
sweet-ann [11.9K]
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3 0
3 years ago
A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
Fantom [35]
In order to accelerate the dragster at a speed v_f = 100 m/s, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
W= K_f - K_i = K_f =  \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is -W_f:
W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W

And since 1 horsepower is equal to 746 W, we can rewrite the power as
P=4.08 \cdot 10^6 W \cdot  \frac{1 hp}{746 W} =547 hp



3 0
4 years ago
Read 2 more answers
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