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3 years ago

Does specific heat of a substance depend on its temperature?

Physics

1 answer:

lara [203]3 years ago

4 0

Answer:

temperature

Explanation:

In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak.

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What is the correct formula for power? A. Power = work / time B. Power = work * time C. Power = force * distance D. Power = work

steposvetlana [31]

Answer:

A. Power = Work / Time

Explanation:

Power is the amount of work done over time, or rather the rate of work, which is given by the unit of watts (W). Since work is defined by Force * Displacement, we can also say Power = Force * Displacement / Time.

3 0

3 years ago

Read 2 more answers

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0

3 years ago

A viscous fluid is flowing through two horizontal pipes. They have the same length, although the radius of one pipe is three tim

Rudik [331]

The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,

$Q_1 = Q_2$

The definition of flow is given by

$Q =VA$

Where

V = Velocity

A = Area

The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.

In this way

$Q_A = Q_B$

$Q_B = 2.94*10^2 m^3/s$

7 0

3 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i

alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

$d=vt+\frac{1}{2}at^2$