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Alex73 [517]
2 years ago
7

Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on ex

pansion, what is its temperature when the initial volume has tripled
Physics
1 answer:
tiny-mole [99]2 years ago
3 0

Answer:

Explanation:

It is a case of adiabatic expansion .

T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}

T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.

Given ,

V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .

( 3 )^{\gamma-1}= \frac{292}{ T_2}

( 3 )^{1.4-1}= \frac{292}{ T_2}

1.552 = 292 / T₂

T₂ = 188 K .

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A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp
Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0
2 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
What are some common characteristics that infectious agents like viruses bacteria fungi and parasites have in common and what ar
CaHeK987 [17]

Answer: Some can and can not kill you

Explanation:

6 0
3 years ago
A woman rides a carnival Ferris wheel at radius 20 m, completing 6.0 turns about its horizontal axis every minute. What are (a)
natta225 [31]

Answer:

Explanation:

Given

radius of ferris wheel is 20 m

It completes 6 turns in 1 min

i.e. 1 turn in 10 sec

Therefore its angular velocity is \omega =\frac{2\pi }{10} rad/s

(a)Period of motion is 10 s

Magnitude of centripetal acceleration is \omega ^2r

=\left ( \frac{\pi }{5}\right )^2\times 20=7.89 m/s^2

(b)Highest point will be 40 m

(c) lowest point 0 m i.e. at ground

7 0
3 years ago
if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for
bagirrra123 [75]

Gravitational force between two masses is given by formula

F = \frac{Gm_1m_2}{r^2}

here we know that

m_1 = 55 kg

m_2 = 1234 kg

r = 3 m

G = 6.67 \times 10^{-11} Nm^2/kg^2

now from the above equation we will have

F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}

F = 5.03 \times 10^{-7}N

so above is the gravitational force between car and the person

5 0
3 years ago
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