The unit of area is called a derived unit.why?

Please fill in the blank if you don't mind.

Liono4ka [1.6K]

Answer:

is there answer choices?

Explanation:

7 0

3 years ago

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Find the heat energy is required to change 2Kg of ice at 0 C to water at 20 C ( specific latent heat of fusion of water = 336000

katrin2010 [14]

We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.

We will find that we must give 840,000 Joules.

First, we must change of phase from ice to water.

We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.

So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:

2*336,000 J = 672,000 J

Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.

Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.

We know that:

specific heat of capacity of water = 4200 J/kg°C

This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.

Then to increase 1°C of 2kg of water we need twice that amount:

2*4,200 J = 8,400 J

And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:

20*8,400 J = 168,000 J

Then the total amount of energy that we must give is:

E = 672,000 J + 168,000 J = 840,000 J

If you want to learn more, you can read:

brainly.com/question/12474790

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3 years ago

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Thepotemich [5.8K]

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

a)

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

$E=KE_i + PE_{si}$

with

$KE_i = 0$ (initial kinetic energy is zero)

$PE_{si} = \frac{1}{2}kx^2$ is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

with

$KE_f = \frac{1}{2}mv^2$ (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

$PE_{sf} = 0$ (elastic potential energy is zero when the spring is released)

Combining the two equations we get

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

And solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s$

b)

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$