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Over [174]
3 years ago
12

consider a sample of gas in a container on a comfortable spring day. the Celsius temperature suddenly doubles, and you transfer

the gas to a container with twice the volume of the first container. If the original pressure was 12 atm, what is a good estimate for the new pressure?
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

6.6 atm

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

Let P₂ be the new pressure

So, P₂ = P₁V₁T₂/V₂T₁

Since V₂ = 2V₁ , P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius

T₂ = 273 + 2t (since its Celsius temperature doubles).

Substituting these values into the equation for P₂, we have

P₂ = P₁V₁(273 + 2t)/2V₁(273 + t)

P₂ = 12(273 + 2t)/[2(273 + t)]

P₂ = 6(273 + 2t)/(273 + t)]

assume t = 30 °C on a comfortable spring day

P₂ = 6(273 + 2(30))/(273 + 30)]

P₂ = 6(273 + 60))/(273 + 30)]

P₂ = 6(333))/(303)]

P₂ = 6.6 atm

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Degger [83]

The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.

<h3>What is induced emf?</h3>

Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.

When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.

The given data in the problem is;

B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T

V(velocity)=125 M/SEC

L(length)=25 cm=0.25 m

The maximum emf is found as;

E=VBLsin90°

E=125 × 5.0 × 10⁻⁵ ×0.25

E=1.56 ×10 ⁻³ V

Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V

To learn more about the induced emf, refer to the link;

brainly.com/question/16764848

#SPJ1

7 0
1 year ago
A 2500-ohm is connected to a 110v power supply. What is the current through the resistor
Assoli18 [71]

Answer:

0.044 amps

Explanation:

Givens

R = 2500 ohms

E = 110 volts

I = ?

Equation

I = E/R

Solution

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7 0
2 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
1 year ago
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