Ask question

Ask question

All categories

3 years ago

15

the pressure difference between an oil and water pipe is measured by double fluid manometer as shown in figure below for the giv

en fluid height and specific gravity calculate the pressure difference

1 answer:

Ahat [919]3 years ago

8 0

The answer above me is correct

You might be interested in

What is dark energy?

Gnesinka [82]

Dark energy is what is causing the accelerating expansion of the universe. It can also be defined as the theoretical energy that counteracts or acts against gravity. Others also define it as a theoretical, new kind of dynamic energy field that fills space, but whose energy has an effect opposite to that of normal energy.

5 0

3 years ago

PLS HELP ME<br> How does one stage in a stars life lead to another

Oxana [17]

Just as during formation, when the material contracts, the temperature and pressure increase. This newly generated heat temporarily counteracts the force of gravity, and the outer layers of the star are now pushed outward ( in not sure tho ) I hope this helps

3 0

3 years ago

Read 2 more answers

A uniform thin rod of length 0.11 m and mass 4.6 kg can rotate in a horizontal plane about a vertical axis through its center. T

ale4655 [162]

Answer:

Explanation:

This problem is based on conservation of rotational momentum.

Moment of inertia of rod about its center

= 1/12 m l² , m is mass of the rod and l is its length .

= 1 / 12 x 4.6 x .11²

I = .004638 kg m²

The angular momentum of the bullet about the center of rod = mvr

where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .

5 x 10⁻³ x v sin60 x .11 x .5 where v is velocity of bullet

According to law of conservation of angular momentum

5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and ( I + mr²) is moment of inertia of bullet rod system .

.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12

.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12

.238 x 10⁻³ v = 55.8375 x 10⁻³

.238 v = 55.8375

v = 234.6 m /s

4 0

3 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago

Which means for obtaining hydrogen from water would require the most energy??

Flauer [41]

I think the correct answer would be to electrolyze water (run an electric current through it) to decompose it into hydrogen and oxygen. Assuming 100% efficiency, it is said that it needs about 40kWh per kilogram of water to fully decompose it.

8 0

3 years ago