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Gre4nikov [31]
3 years ago
7

A diver 50 m deep in 10∘C fresh water exhales a 1.0-cm-diameter bubble. What is the bubble's diameter just as it reaches the sur

face of the lake, where the water temperature is 20∘C?
Physics
1 answer:
aliya0001 [1]3 years ago
4 0

Answer:

18.2mm

Explanation:

D = 50m

T1 = 10+273 = 283K

T2 = 20+273 = 293K

R1 = 5x10^-3

Absolute pressure at 50m

P1 = pA + pwateer x g x d

= 101000+ 1000x9.81x50

= 591500pa

New volume of bubble

= P1v1/T1 = p2v2/T2

= 125x10^-9 x 591500x293/101000*283

= 757.9x10^-9m³

R2 = 9.2x10^-3

D2 = 18.2mm

Or 1.82cm

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a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be
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The equivalent resistance of n resistors connected in parallel is given by
\frac{1}{R_{eq}} =  \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n} (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is R_{Eq}=2.00 \Omega, and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R}   + \frac{1}{R}   + \frac{1}{R}   + \frac{1}{R}   =  \frac{5}{R}
From which we find R= 5 \cdot 2 \Omega = 10 \Omega.

So, each piece of wire has a resistance of 10 \Omega. Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
R_{Eq}=R_1 + R_2 + ...+R_n
So if we apply it at our case, we have
R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega

therefore, the resistance of the original wire was 50 \Omega.
5 0
3 years ago
A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2
Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

4 0
3 years ago
Which characteristic of a light wave must increase
anzhelika [568]
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ivanzaharov [21]

Answer:

4A

Explanation:

According to ohm's law;

E = IRt where;

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I is the total current flowing in the circuit = ?

Rt is the total effective resistance in the circuit.

To find Rt, we will resolve the resistors in parallel first.

Since 6ohms and 12ohms resistors are in parallel, their effective resistance will give;

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1/R = 3/12

3R = 12

R = 4ohms.

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Rt = 4+2

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I = 24/6

I = 4Amperes

The total current in the circuit is 4A

This same currents will flow in the 2ohms resistor since same current flows in a series connected resistors.

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4 years ago
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