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3 years ago

En un dia de invierno en la cd de paris se registró una temperatura de -10°c convertir ese temperatura a °k

Physics

1 answer:

schepotkina [342]3 years ago

6 0

Answer:

-10°C = 263 Kelvin

Explanation:

Given the following data;

Temperature in °C = -10

To convert the value of the temperature to Kelvin;

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object. It is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

Next, we would convert the temperature in degree Celsius to Kelvin by using the following mathematical expression;

Kelvin = 273 + °C

Substituting into the formula, we have;

Kelvin = 273 + (-10)

Kelvin = 273 - 10

Kelvin = 263 K

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3 years ago

A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.

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Answer:

Correct answer: Third statement P = 4900 W

Explanation:

Given:

m = 500 kg the mass of the elevator

h = 10 m reached height after t = 10 seconds

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P = W / t

since work is a measure of change in this case of potential energy then it is:

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4 years ago

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3 years ago

4-12. The morning inspection of the tank farm finds a leak in the turpentine tank. The leak is repaired. An investigation finds

bonufazy [111]

Answer:

a) $V=759.8727\ ft^3$

b) $\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}$

c) $t\approx29.541\ s$

Explanation:

Given:

diameter of hole in the tank, $d'=0.1\ in=\frac{1}{120}\ ft$

position of the hole form the tank bottom, $h' =7\ ft$

initial level of turpentine in the tank before the leakage, $h_i=17.3\ ft$

level of turpentine in the tank after the repair of leakage, $h_f=13\ ft$

diameter of the tank, $d=15\ ft$

density of turpentine oil, $\rho=55\ lbm.ft^3$

Now, volume of turpentine spilled:

$V=A.(h_i-h_f)$

where:

$A=$ area of the cross section of the tank's volume

$V=\pi .\frac{d^2}{4} \times(h_i-h_f)$

$V=\pi\times\frac{15^2}{4} \times(17.3-13)$

$V=759.8727\ ft^3$

When the tank was full the liquid level was highest:

so velocity form the height of the hole will be given as:

$v=\sqrt{2g.(h_i-h')}$

$v=\sqrt{2\times 32.12\times (17.3-7) }$

$v=25.722\ ft.s^{-1}$

Now we have the flow rate of the spillage given by:

$\dot V=(\pi.\frac{d'^2}{4}) \times v$

$\dot V=\pi\times \frac{(\frac{1}{120})^2 }{4} \times 25.722$

$\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}$

Total time the leak was active can be calculated as:

$t=\frac{V}{\dot V}$

$t=\frac{759.8727}{25.722}$

$t\approx29.541\ s$

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