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kow [346]
3 years ago
14

Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab

(volume of 180m3) was 1.1 atm before the storm and 0.85 atm during the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
a. 190m3
b. 230m3
c. 1,800m3
d. 7,100m3
Physics
1 answer:
forsale [732]3 years ago
7 0

Answer:

232.9m³ (Option b. is the closest answer)

Explanation:

Given:

Air pressure in the lab before the storm, P₁ = 1.1atm

Air volume in the lab before the storm, V₁ = 180m³

Air pressure in the lab during the storm P₂ = 0.85atm

Air volume in the lab before the storm, V₂ = ?

Applying Boyle's law:    P₁V₁ = P₂V₂    (at constant temperature)

                          V_{2} = \frac{P_{1}V_{1}}{P_{2}}

                          V_{2} = \frac{1.1 * 180}{0.85}

                          V_{2} = \frac{198}{0.85}

                           V₂  = 232.9m³

The air volume in the laboratory that would expand in order to make up for the large pressure difference outside is 232.9m³

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A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil cont
EastWind [94]

Answer:

The number of turns in secondary coil is 4000

Explanation:

Given:

Current in primary coil I_{P} = 500 A

Current in secondary coil I_{S} = 25 A

Number of turns in primary coil N_{P} = 200

In case of transformer the relation between current and number of turns is given by,

     \frac{N_{S} }{N_{P}  } = \frac{I_{P} }{I_{S} }

For finding number of turns in secondary coil,

     N_{S} = \frac{I_{P} }{I_{S} }  N_{P}

     N_{S} = \frac{500}{25} \times 200

     N_{S} = 4000

Therefore, the number of turns in secondary coil is 4000

5 0
3 years ago
The atmosphere is held together by
Vera_Pavlovna [14]

Answer:

D. gravity

Explanation:

Gravity keeps the atmosphere from escaping into space.

3 0
4 years ago
Why does it take much longer time and distance to stop a moving ship in water than a moving car on the road
Naddik [55]
The force on the ship is more than a car
6 0
3 years ago
You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
3 years ago
3.7. A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle 23.1 degrees north of east, and finally 15.0 m west.
andriy [413]

Answer:

Explanation:

We shall represent displacement of dog in vector form , in terms of i , j , i representing east  and  j representing north .

Dog travels 3.5 m south .

Displacement D₁ = - 3.5 j

then dog travels 8.2 m , 23.1 degree north of east

Displacement D₂ = 8.2 cos23.1 i + 8.2 sin23 j

D₂ = 8.2 cos23.1 i + 8.2 sin23.1  j

= 7.54 i + 3.22 j  

Third displacement

D₃ = - 15i

Total displacement = D₁ + D₂ + D₃

= - 3.5 j + 7.54 i + 3.22 j  -15i

= - 7.46 i - 0.28 j

Magnitude of displacement = √ ( 7.46² + .28²)

= √(55.65 + .08 )

= 7.46 m

b ) Direction of displacement

If Ф be angle , displacement makes with west direction

TanФ =  .08 / 55.65 = .00143

Ф = .082 degree south of west or almost west .

From east , this angle = 180 + .082 = 180.082 , counterclockwise .

5 0
3 years ago
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