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2 years ago

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

Physics

1 answer:

Lady bird [3.3K]2 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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Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?

Ede4ka [16]

Answer:

W=f×d

w=45×27

w=1215 j.............

3 0

3 years ago

Water ﬂows through a horizontal nozzle in steady ﬂow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2

Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

Q = A v

v = Q / A

The area of a circle is

A = π r² = π d² / 4

Let's look at the speeds at each point

v₁ = Q / A₁ = Q 4 /π d₁²

v₁ = 10 4 /π 0.5²

v₁ = 50.93 m / s

v₂ = Q / A₂

v₂ = 10 4 /π 0.25²

v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

P₁ = P2 + ½ rho (v₂² - v₁²)

P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

P₁ = 1.013 10⁵ + 2.205 10⁷

P₁ = 2.215 10⁷ Pa

la definicion de presion es

P₁ = F₁/A₁

F₁ = P₁ A₁

F₁ = 2.215 10⁷ pi d₁²/4

F₁ = 2.215 10⁷ pi 0.5²/4

F₁ = 4.3 106 N

6 0

2 years ago

A bulb pile is driven to the ground with a 2.5 ton hammer. The drop height is 22 ft and the volume in last batch driven is 4 cu

Stels [109]

Answer:

159.1 ton

Explanation:

The solution is shown in the attached file

Download docx

5 0

3 years ago

Read 2 more answers

It is easier to climb up a slanted slope than a vertical slope

V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

3 0

2 years ago

Read 2 more answers

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago