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Anna35 [415]
3 years ago
13

A 20 liter cylinder of helium at a pressure of 150 atm and a temperature of 27ÁC is used to fill a balloon at 1.00 atm and 37ÁC.

What is the volume of the balloon?
A) 2400 Liters

B) 2900 Liters

C) 3000 Liters

D) 3100 Liters
Physics
2 answers:
djyliett [7]3 years ago
7 0

<u>Answer</u>

D) 3100 Liters


<u>Explanation</u>

To get the volume if the balloon you need to use the combined equation of the low of gases.

P₁V₁/T₁ = P₂V₂/T₂

(20×150)/(27+273) = (1×V₂)/(37+273)

3000/300 = V₂/310

10 = V₂/310

V₂ = 10 × 310

    = 3100 Liters


LekaFEV [45]3 years ago
4 0
Using the ideal gas relationship, we know that:
PV/T = constant
For a fixed amount of gas.

P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/T₁P₂
V₂ = (150 x 20 x (27 + 273)) / ((37 + 273) x 1)
V₂ = 2903 L
= 2900 L
The answer is B
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satela [25.4K]

Answer:

50 kg

Explanation:

Given,

Force ( F ) = 100 N

Acceleration ( a ) = 2 m/s^2

To find : Mass ( m ) = ?

Formula : -

F = ma

m = F / a

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m = 50 kg

Therefore, the mass of the object is 50 kg.

5 0
3 years ago
Hydraulic press is called an instrument for multiplication of force. Why?
Lisa [10]

Answer:

Hydraulic press is called an instrument for multiplication of force. Why? Because it uses Pascal's idea and  principle: F=p*S. If we apply small force to small piston you will generate a pressure. According to Pascal's law pressure is the same everywhere in closed system so the same pressure will act on large piston on the other side too.

Explanation:

4 0
3 years ago
The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single
Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W =  mg = (689/1000)9.8 = 6.75 N,  D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0
3 years ago
A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads
Bond [772]

The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is calculated as follows;

Capacitors in series;

1/Ct = 1/8 + 1/7.5

1/Ct = 0.25833

Ct = 3.87 mF

Capacitors is parallel;

Ct = 3.87 mF + 12 mF + 15 mF

Ct = 30.87 mF

Ct = 0.03087 F

<h3>Charge in each capacitor</h3>

Q = CV

Q = 0.03087 x 12

Q = 0.37 C

Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

Learn more about capacitors here:  brainly.com/question/13578522

#SPJ1

3 0
2 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi
Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

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                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

                      a = -\mu g

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              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0
3 years ago
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