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professor190 [17]
4 years ago
6

A small bubble rises from the bottom of a lake, where the temperature and pressure are 4°C and 3.0 atm, to the water's surface,

where the temperature is 25°C and the pressure is 0.95 atm. Calculate the final volume of the bubble if its initial volume was 2.1 mL.
Physics
1 answer:
qwelly [4]4 years ago
6 0

Answer:

7.13mL

Explanation:

P₁V₁T₁ = P₂V₂T₂

P₁ = 3atm , V₁ = 2.1 mL , T₁ = 273 + 4 =277K

P₂ = 0.95atm , V₂ = ? , T₂ = 273 + 25 =298K

V₂ = P₁V₁T₂ / P₂T₁

V₂ = (3atm)(2.1 mL )(298K) / (0.95atm)(277K)

V₂ = 7.13mL

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20 points plz help i dont really understand how to do this
Artist 52 [7]

\huge{ \mathfrak{  \underline{ Answer} \:  \:  ✓ }}

1. Calculate the equivalent resistance in the circuit :

In the given diagram, the resistors are in series connection,

So equivalent resistance = sum of resistance of both resistors.

\large \boxed{R_{eq} =R_1 +R_2  }

  • R_{eq} =  480 + 360

  • R_{eq} = 840 \: ohms

Therefore, Equivalent resistance = 840 ohms.

2. Calculate the current through the battery :

\boxed{ \mathrm{current =  \dfrac{potential \:  \: difference}{resistance} }}

  • i =  \dfrac{120}{840}

  • i =  \dfrac{1}{7}

  • i = 0.142 \: Amperes

Hence, current through the battery = 0.142 A

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___________________________

\mathrm{ \#TeeNForeveR}

4 0
3 years ago
A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30
Ivan

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c)  1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed  when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis )  ----- ( 1 )

f ( friction force )=  ( 16400v * 6.25 *10^-2) =  1025 v

F ( force applied ) = 75

W ( weight of  block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = \frac{75- (98.1*sin30^{0}) }{1025} = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = \frac{0 - (98.1*sin30^{0}) }{1025} =  - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = \frac{75+( 98.1*sin30^{0})  }{1025} = 1.21 * 10^-1 m/s

5 0
3 years ago
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