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kolbaska11 [484]
3 years ago
13

numerical question : what is the required heat to raise the temperature of 2 kg parrafin by 10 Celsius if 44000 joules is requir

ed to raise it by 200 celsius​
Physics
1 answer:
kirill115 [55]3 years ago
8 0

Answer:

The heat energy required is, E = 2200 J

Explanation:

Given,

The mass of paraffin, m = 2 Kg

The energy required to raise the temperature of the paraffin by 200° C = 44000 J

Then the heat energy required to raise the temperature of the paraffin by 10° C is given by,

Since 44000 J raises temperature by 200° C, then

                              E = 44000 J / 20

                                 = 2200 J

Hence, the energy required to raise the temperature of the paraffin by 10° C is, E = 2200 J

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Answer:

A

Explanation:

There are three basic forces in aerodynamics: acceleration, which moves an airplane forward; drag, which holds it back; and height, which keeps it airborne. Lift is generally explained by three theories: Bernoulli's principle, the Coanda effect, and Newton's third law of motion.

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Driving along a boring stretch of interstate in Illinois, you start experimenting using the average speed equation you learned i
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The average speed would be 33.29m/s.
The average speed equation is:

Average speed =  \frac{total distance}{total time}

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled. 

Scenario 1:
Speed = 29m/s
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Scenario 2
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Distance = (35m/s)(300s)
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What physical quantity is a measure of the amount of inertia an object has?
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A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
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Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

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now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

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