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never [62]
1 year ago
7

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

erature at 27 °C. What will be the new volume if the balloon is when it moves upwards to where the atmospheric pressure is 40 kPa and the temperature is at 10 °C?​
Physics
1 answer:
Yanka [14]1 year ago
7 0

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

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In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.
garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...

Find the equivalent resistance of the right branch of the circuit:

R_{eq}  = R_{2} +R_{3} \\R_{eq} = 20.7 + 25.8 = 46.5 ohms

\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411

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2 years ago
In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

brainly.com/question/2562955

brainly.com/question/9575487

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#LearnwithBrainly

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3 years ago
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zvonat [6]

Answer:

dry air is way heaver

Explanation:

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yarga [219]
8 miles per hour

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Which quantities indicate a direction and a magnitude?<br>Check all that apply.​
Korvikt [17]

Answer:

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Explanation:

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3 years ago
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