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Bad White [126]
3 years ago
15

If you know the power rating of an appliance and the voltage of the line it is attracted to , you can calculate the current the

appliance uses by
Physics
2 answers:
marshall27 [118]3 years ago
8 0

                                Power = (voltage) times (current)

Divide each side
by voltage:              Power / voltage  =  current .

If you know the power and the voltage, then you can divide
the power by the voltage, and the quotient is the current.
rusak2 [61]3 years ago
3 0

Dividing The Power by the voltage im pretty sure.

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A piece of aluminium foil is held between a bar magnet and a paperclip. What will happen to the paper clip?
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Answer:the paper clip will be attracted to the magnet

Explanation:

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The temperature of a substance during an experiment can be modeled by the function f(x)=−7.5cos(πx30)+31.4 , where f(x) is the t
vekshin1
38.9 degrees celsius
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Match these items.
aleksandrvk [35]

Answer:

Here's your answer :

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hope it helps!

3 0
3 years ago
A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o
IgorC [24]

Answer:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

Explanation:

For this case we know the mass of the water given :

m = 787 gr

And we know that the initial temperature for this water is T_i =75 C.

We want to cool this water to the human body temperature T_f = 37 C

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Q= m c_p \Delta T

Where c_p represent the specific heat for the water and this value from tables we know that c_p =1 \frac{cal}{gr C} for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

8 0
3 years ago
Read 2 more answers
All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outle
VARVARA [1.3K]

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage   ......................1

here voltage is same in all three device

so

current by toaster is

I = \frac{1140}{120}

I = 9.5 A

and

current by blender

I = \frac{270}{120}

I = 2.25 A

and

current by lamp is

I = \frac{80}{120}

I = 0.667 A

so here device in parallel so

total current is = 9.5 A + 2.25 A + 0.667 A

total current = 12.417 A

so it will not fuse as current is less than 15 A

8 0
3 years ago
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