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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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PLZZZZ HELP QUICK . 30 points. Emma weighs 560 N, she has decided to stay in shape so she is working out every morning, This mor
Phantasy [73]

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

3 0
2 years ago
Which image illustrates refraction?
maks197457 [2]

Answer:

B illustrates refraction

3 0
3 years ago
Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box
photoshop1234 [79]

Answer:

Bill's motor power: W_B = F x S / T = F x 0.35 / 2= 0.175F

Nageen's motor power: W_N = F x S / T = F x 0.35 / 1.8 = 0.194F

=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.

5 0
3 years ago
What is the wavelength of radar waves with the frequency of 6.0 x 10 ^10 Hz?
Serjik [45]

Yes yes multiply hurry up

8 0
3 years ago
The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
3 years ago
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