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2 years ago

At what water temperature will additional heat energy need to be added before the temperature will change again?

1 answer:

7 0

That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)

Hope this helps! (If correct, please rank as brainliest answer) :)

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If i apply 280 n of force to a 40kg object, what will it's acceleration be?

nika2105 [10]

Answer:

f=ma

f=280N

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a=?

280=40a

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2 years ago

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder

beks73 [17]

Answer:

$d=1.67mi$

Explanation:

Assuming the light takes essentially no time to reach you, the distance at which the lightning occurred can be calculated by multiplying the speed of sound by the time it takes to hear the thunder:

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2 years ago

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Answer:

The answer is D

Explanation:

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3 years ago

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A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence

Anestetic [448]

Answer:

The linear velocity is $v=4.08m/s$

Explanation:

According to the law of conservation of energy

The potential energy possessed by the hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

The kinetic energy can be mathematically represented as

$KE = \frac{mv^2}{2} + \frac{Iw}{2}$

Where $I$ is the moment of inertia possessed by the hoop which is mathematically represented as

$I = mr^2$

Here R is the radius of the hoop

$w$ is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

$w = \frac{v}{r}$

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

Now expressing the above statement mathematically

$potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}$

$mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}$

=> $mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}$

=> $mgh = \frac{mv^2}{2} + \frac{mv^2}{2}$

=> $mgh = mv^2$

=> $v = \sqrt{gh}$

Substituting values

$v = \sqrt{9.81 * 1.7}$

$v=4.08m/s$

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2 years ago

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What type of lens does a camera contain?

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