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2 years ago

What happens to the density of a fluid as it temperature increases/decreases?

1 answer:

4 0

Let's look at the density of water at 25 deg C and compare that to a higher temperature, 80 deg C. The density decreases from 0.9970 g/mL to 0.9718 as it is heated. This makes sense because, as heat is added to the liquid water, there is greater kinetic energy of the molecules and there are also more vibrations of the water molecules. Together these mean that each H2O unit in liquid water takes up more space as the temperature increases.

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A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferre

Paraphin [41]

Answer: 80J

Explanation:

According to the first principle of thermodynamics:

<em>"Energy is not created, nor destroyed, but it is conserved." </em>

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy $U$ , which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

$\Delta U=Q+W$ (1)

Where:

$\Delta U$ is the variation in the internal (thermal) energy of the system (the value we want to find)

$Q=-100J$ is the heat transferred out of the gas (that is why it is negative)

$W$ is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

$W=-P \Delta V$ (2)

Where:

$P=450kPa=450(10)^{3}Pa$ is the constant pressure of the gas

$\Delta V=V_{f}-V_{i}$ is the variation in volume of the gas

In this case the initial volume is $V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}$ and the final volume is $V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}$ .

This means:

$\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}$ (3)

Substituting (3) in (2):

$W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})$ (4)

$W=180J$ (5)

Substituting (5) in (1):

$\Delta U=-100J+180J$ (6)

Finally:

$\Delta U=80J$ This is the change in thermal energy in the compression process.

8 0

3 years ago

A 20 m high filled water tank develops a 0.50 cm hole in the vertical wall near the base. With what speed does the water shoot o

Rina8888 [55]

Answer:

The speed of the water shoot out of the hole is 20 m/s.

(d) is correct option.

Explanation:

Given that,

Height = 20 m

We need to calculate the velocity

Using formula Bernoulli equation

$\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}$

Where,

v₁= initial velocity

v₂=final velocity

h₁=total height

h₂=height of the hole from the base

Put the value into the formula

$v_{1}^2=2g(h_{2}-h_{1})$

$v_{1}=\sqrt{2g(h_{2}-h_{1})}$

$v_{1}=\sqrt{2\times9.8\times(20-0.005)}$

$v_{1}=19.7\ m/s= approximate\ 20\ m/s$

Hence, The speed of the water shoot out of the hole is 20 m/s.

7 0

3 years ago

Las carreras de velocidad pura en el atletismo son:

zheka24 [161]

La respuesta es la letra b

8 0

3 years ago

Consider the minute hand on a clock. (a) Compute the frequency of its motion in cycles per second. State your answer to three si

kicyunya [14]

Answer:

(a)0.0002778

(b) $1.16\times10^{-5}$

Explanation:

(a) The minute hand has a period of 60 minutes ( or 60 * 60 = 3600 seconds) for 1 circle. Its frequency per second would be

1 / 3600 = 0.0002778

(b) The hour hand has a period of 24 hours ( or 24*60 * 60 = 86400

seconds) for 1 circle. Its frequency per second would be

$1 / 86400 = 1.16\times10^{-5}$

5 0

3 years ago

Hello help me pls! i need serious help

9966 [12]

Answer:c

Explanation:

7 0

2 years ago

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