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Gennadij [26K]
3 years ago
14

Consider two soap bubbles having radii r1 and r2 (r1 < r2) connected via a valve. What happens if we open the valve?

Physics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

Explanation:

The radius of the smaller bubble, r1 will decrease and that of the bigger bubble, r2 will increase.

The pressure that is present in the smaller bubble usually is greater than the pressure that exists inside that of the bigger bubble. This then makes air to flow from r1 to r2 thereby making the radius of the smaller bubble r1, to decrease while keeping that of the bigger bubble r2 higher.

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When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of
Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0
4 years ago
A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of
Leto [7]

Answer: 2.27\ m/s^2

Explanation:

Given

Length of the race track L=200\ m

the radius of curvature of the track r=29.5\ m

time taken to run on track is t=24.4\ s

Speed of runner is

\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s

Centripetal acceleration is

\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2

5 0
3 years ago
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed
jenyasd209 [6]

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current  I  = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41  B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

7 0
3 years ago
An object accelerates to a velocity of 34m/s over a time of 1.3 s the acceleration it experienced was 15 m/s squared what was th
just olya [345]
V=v0+at=>v0=v-at=34-15•1,3=34-19,5=14,5m/s
7 0
3 years ago
Read 2 more answers
Select all the correct answers.
WARRIOR [948]
Oop i have no clue but i need to answer questions so that i can ask some :)
3 0
3 years ago
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