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Vadim26 [7]
3 years ago
8

What would cause the pressure in a sealed container of gas to increase?

Physics
1 answer:
Whitepunk [10]3 years ago
3 0
<span>a. The temperature increases.</span>
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Referring to the graph below of data describing a cart pulled across a level surface. When a 4 N force pulls on the cart, the ca
LenKa [72]

Answer:gjv

Explanation:

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sveticcg [70]

Answer:

I believe the answer is A and D.

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A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere
dmitriy555 [2]

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

5 0
3 years ago
If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will
densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

P=\frac{V^2}{R}

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P

so, the power dissipated will quadruple.

6 0
3 years ago
A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field
Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= -  .086 Weber

change in flux

.15 +  .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0
3 years ago
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