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2 years ago

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calculate the pressure exerted by a mercury column of 76 cm high at its bottom.Given that density of mercury is 13600kg/m^3 and

g=9.8

1 answer:

Amanda [17]2 years ago

5 0

Answer:

Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me ...

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What two particles in an atom are equal in number

Ainat [17]

The two particles that are equal in a atom is the proton and electron.

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3 years ago

What does K stand for in Hookes law?

wlad13 [49]

So Hooke's law says that that law is proportional to how much I stretch the spring. Alright. So f=kx<span>. x is the length of the spring now minus its length when it's relaxed and nobody's pulling on it. k is a constant called the spring constant.</span>

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3 years ago

Read 2 more answers

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

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3 years ago

Convert 463.833 m/s to km/ min

Alexus [3.1K]

Answer:

27.82998 km/min

Explanation:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

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2 years ago

Uploaded imageTwo long, parallel wires each carry the same current I in the same direction (see figure below). The total magneti

LekaFEV [45]

Answer:

Zero

Explanation:

Two long parallel wires each carry the same current I in the same direction. The magnetic field in wire 1 is given by :

$B_1=\dfrac{\mu_oI_1}{2\pi r}$

Magnetic force acting in wire 2 due to 1 is given by :

$F_{2}=I_2lB_1$

$F_2=\dfrac{\mu_oI_1I_2 l}{2\pi r}$

Similarly, force acting in wire 1 is given by :

$F_1=\dfrac{\mu_oI_1I_2 l}{2\pi r}$ According to third law of motion, the force acting in wire 1 will be in opposite direction to wire 2 as :

$F_2=-F_1$

So, the total magnetic field at the point P midway between the wires is in what direction will be zero as the the direction of forces are in opposite direction.

7 0

2 years ago