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3 years ago

What is the atmospheric pressure and temperature at sea level in a standard atmosphere?

Physics

1 answer:

Lady bird [3.3K]3 years ago

4 0

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

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How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?

Illusion [34]

Power is defined as the rate at which the body is doing work:
$P=\frac{W}{t}$
Work is defined as displacement done by the force times that displacement:
$W=F\cdot h$
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
$W=62N\cdot10m=620J$
of work.
Now we just divide that by 5s to get how much power is required:
$P=\frac{620J}{5s}=124W$

5 0

3 years ago

Why is venus called brightest star and mars called red planet

bogdanovich [222]

Answer:given below by him is correct

Explanation:

Pls refer his

3 0

3 years ago

Find the emitted power per square meter of peak intensity for a 3000 k object that emits thermal radiation.

Aleksandr [31]

According to Stefan-Boltzmann Law, the thermal energy radiated by a radiator per second per unit area is proportional to the fourth power of the absolute temperature. It is given by;
P/A = σ T⁴ j/m²s
Where; P is the power, A is the area in square Meters, T is temperature in kelvin and σ is the Stefan-Boltzmann constant, ( 5.67 × 10^-8 watt/m²K⁴)
Therefore;
Power/square meter = (5.67 × 10^-8) × (3000)⁴
= 4.59 × 10^6 Watts/square meter

8 0

3 years ago

Read 2 more answers

Calculate 2 x 10^-3cm ÷ 2.5 x 10^4cm

Mamont248 [21]

Here is your answer:

First find the notations:

2×10^-3
=0002
And...
2.5×10^4=25000

Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8

3 0

3 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

Svetllana [295]

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

L = 2.7x10^32W, T = 11,000K, s= 5.67 x 10^-8, R= radius in meters

Rigel parallax, p = 0.00378 arc sec

Substituting the various values and making R the subject of the formula

R² = L/(4psT⁴)

R² = 2.7x10^32/(4 x 0.003878 x 5.67x10^-8 x (11,000)⁴)

R² = 2.7x10^32/1.2877x10^7

R² = 2.096761668 x 10^25

R = 4.579041021 x 10^12meters

(b)

Procyon B, 2.1x10^23W, T = 10,000K

But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

R = 1.80565 x 10^8 meters

(c) The radius of Rigel is given as 4.579041021 x 10^12meters and the radius of Procyon B is given by 1.80565 x 10^8 meters shows the remarkable difference between a super-giant star(Rigel) and a white dwarf star (Procyon B)

The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

The sun is larger than Procyon B as estimated above.

7 0

3 years ago