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PtichkaEL [24]
3 years ago
10

The amount of power required to move an object can be increased without changing the

Physics
1 answer:
VLD [36.1K]3 years ago
3 0

Answer:

Yes it is possible to increase the power with out changing the amount of work.

Explanation:

The power is defined by the amount of power divided by the time. This time is the one needed to do the work. We can understand this issue by analyzing an example with numeric values.

Work = 500 [J]

Time = 5 [s]

Power will be:

Power=\frac{500}{5} \\Power=100 []watt]\\

Now if we change the time to 2 seconds:

Power = 500 [J]/2[s]\\Power = 250 [watt]\\

As we can see, the power was increased without the need to change the work.

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3 years ago
A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc
Lana71 [14]

1. - 12N

2.- 2 hours

3. - 0.5 m/s/s

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3 years ago
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suppose that you had an ohmmeter and a mystery component that could be either an inductor or a capacitor. How could you use the
sukhopar [10]

Answer:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

if the resistance > 1( very high )   then component is a capacitor

if the resistance = 0 then component is an inductor

Explanation:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

if the resistance > 1( very high )   then component is a capacitor

if the resistance = 0 then component is an inductor

8 0
2 years ago
In which medium does sound travel the fastest?
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Sound travels through solids the fastest

Hope it helps...

7 0
3 years ago
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An inventor claims to have developed a power cycle having a thermal efficiency of 40%, while operating between hot and cold rese
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From Carnot's theorem, for any engine working between these two temperatures:


efficiency <= (1-tc/th) * 100


Given: tc = 300k (from question assuming it is not 5300 as it seems)

For a, th = 900k, efficiency = (1-300/900) = 70%

For b, th = 500k, efficiency = (1-300/500) = 40% 

For c, th = 375k, efficiency = (1-300/375) = 20% 


Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid. 

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3 years ago
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