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diamong [38]
3 years ago
9

What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure

Physics
1 answer:
mezya [45]3 years ago
8 0

Answer:

this is a no brainer

Explanation:

As air pressure in an area increases, the density of the gas particles in that area increases.

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Afina-wow [57]

Answer:

For A

Displacement= 1/2*3*6= 9m

For B

Displacement= 1/2*4*4= 8m

5 0
3 years ago
The element in an incandescent light bulb that releases light energy is
Leviafan [203]
A thin tungsten filament
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3 years ago
What are the top two gasses in the earth’s atmosphere ?
ch4aika [34]

Answer:

Nitrogen and oxygen are by far the most common; dry air is composed of about 78% nitrogen (N2) and about 21% oxygen (O2). Argon, carbon dioxide (CO2), and many other gases are also present in much lower amounts; each makes up less than 1% of the atmosphere's mixture of gases.

7 0
3 years ago
2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?
yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0
3 years ago
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
3 years ago
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