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olasank [31]
3 years ago
11

An office window has dimensions 3.7 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.972 a

tm, but inside the pressure is held at 1.0 atm. What net force pushes out on the window?
Physics
1 answer:
Katyanochek1 [597]3 years ago
8 0

Answer:

22044.3 N

Explanation:

We are given that

Dimension of office window=3.7m\times 2.1 m

Area of office window=l\times b=3.7\times 2.1=7.77 m^2

Outside air pressure=P1=0.972 atm

Inside pressure,P2=1.0 atm

Pressure=P2-P1=1-0.972=0.028 atm=0.028\times 101325 Pa=2837.1 Pa

1 atm=101325 Pa

Force,F=PA

Using the formula

F=7.77\times 2837.1=22044.3 N

Hence, the net force pushes out on the window=22044.3 N

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Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di
vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

A = \pi (\frac{D}{2})^2

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

d = \pi r^2 e

r^2_e= \frac{d}{\pi}

r_e = \sqrt{\frac{d}{\pi} }

replacing d = \frac{\pi (\frac{D}{2})^2 }{10, 000} in the equation above; we have:

r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }

r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}

r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}

r_e = 500 ly

The distance (s) between each civilization = 2(r_e)

= 2 (500 ly)

= 1000 light-years (ly)

4 0
3 years ago
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
I'm feeling nice so.... F.r.e.e POINTS (pls follow me) ​
NISA [10]

Answer:

ty :,)

Explanation:

6 0
3 years ago
Read 2 more answers
What is a substance
Korolek [52]

Explanation:

A particular kind of matter with uniform properties..

8 0
3 years ago
Read 2 more answers
Which represents the net force
ipn [44]

Answer:

The net force is 500N downwards

Explanation:

When Haley is trying to pull an object upward. The below forces are acting on the object.  

Fp = 5500N

Fg = 6000N

because the force of gravity is more than the force of the pull.

Fnet = Fg - Fp = 6000N - 5500N = 500N

And, the direction of the resultant force is the direction of the larger force.

4 0
3 years ago
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