# A hair dryer uses 75,000 joules of energy in 60 seconds. What is the power of this hair dryer?

Power is the amount of energy transferred in a second or, to put it more simply, the rate of energy transfer with time. In your case:

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The power of the hair dryer can be calculated using the formula:

[ \text{Power} = \frac{\text{Energy}}{\text{Time}} ]

Given:

- Energy used by the hair dryer (E) = 75,000 joules
- Time taken (t) = 60 seconds

Using the formula:

[ \text{Power} = \frac{75,000 , \text{joules}}{60 , \text{seconds}} ]

[ \text{Power} = 1250 , \text{watts} ]

Therefore, the power of the hair dryer is 1250 watts.

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To calculate the power of the hair dryer, you can use the formula:

Power (P) = Energy (E) / Time (t)

Given that the hair dryer uses 75,000 joules of energy in 60 seconds:

Power (P) = 75,000 joules / 60 seconds

Now, you can calculate the power:

P = 1250 watts

Therefore, the power of this hair dryer is 1250 watts.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- If a #2 kg# object is constantly accelerated from #0m/s# to #16 m/s# over 6 s, how much power musy be applied at #t=3 #?
- A ball with a mass of #250 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #8 (kg)/s^2# and was compressed by #3/2 m# when the ball was released. How high will the ball go?
- A ball with a mass of #70 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #5 (kg)/s^2# and was compressed by #7/6 m# when the ball was released. How high will the ball go?

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