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Brut [27]
3 years ago
10

Radon is a radioactive gas found below ground. If a sample of radon occupies 29 mL at 25 oC, what volume will it occupy at 500 K

? Assume constant pressure.
Physics
1 answer:
Valentin [98]3 years ago
8 0

Answer:

48.63 mL

Explanation:

Using Charles's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 29 mL

V₂ = ?

T₁ = 25 °C

T₂ = 500 K

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using above equation as:

\frac{29\ mL}{298.15\ K
}=\frac{V_2}{500\ K}

V_2=\frac{29\cdot \:500}{298.15}\ mL

<u>New volume = 48.63 mL</u>

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The nucleus of an atom can have either a positive or negative charge. please select the best answer from the choices provided t
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F - False.

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if the mass is 10 gram and the volume is 7 cubic centimetre,find the density in kilogram per cubic metre
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1.42

Explanation:

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7 0
3 years ago
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p
Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

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So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0
2 years ago
How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?​
grandymaker [24]

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

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3 years ago
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