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arlik [135]
2 years ago
10

Why is the temperature of the liquid in the flask on the previous page measured when the liquid in the thermometer has stopped r

ising? Explain
How can the thermometer in the flask on the previous page be used to demonstrate the relationship between heat transfer and kinetic energy? Explain.
Physics
1 answer:
Flura [38]2 years ago
3 0

when heated, the molecules of the liquid in the thermometer move faster, causing them to get a little further apart. this results in movement up the thermometer. when cooled, the molecules of the liquid in the thermometer move slower, causing them to get a little closer together.

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Two positive charges of same magnitude are separated by some distance if we bring a unit positive charge from one charge to anot
Studentka2010 [4]

Answer:

Increases.

Explanation:

The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.

3 0
3 years ago
What are two reasons why a home computer scanner requires
alexandr1967 [171]

Answer: C and D

Explanation: a p e x

4 0
2 years ago
Read 2 more answers
.A particle starts on the origin. It is pushed back to -5.7 m in 2.1 s. Then it is pushed
harina [27]

Answer: The average velocity is -0.965m/s

Explanation: The first step is to calculate the two velocities is both directions. A velocity is a distance per unit time.

V=d/ t

=-5.7/2.1

=-2.7m/s

For the other direction the velocity is

V=7.3/9.5

=0.77m/s

The average velocity the add the velocities and divide them by 2.

V=-2.7+0.77/2

V= 0.965m/s

5 0
3 years ago
Read 2 more answers
A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a c
tester [92]

Answer:

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

Explanation:

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.

3 0
3 years ago
A baseball of radius r = 5.2 cm is at room temperature T = 20.8 C. The baseball has emissivity of ε = 0.86 and the Stefan-Boltzm
Marrrta [24]

Answer:

P = 12.37 \frac{J}{s} = 12.37 Watts

Explanation:

Previous concepts

The Thermal radiation is one of "3 mechanisms who allows to bodies exchange energy".

The thermal radiation formula is given by:

\frac{P}{A} = \epsilon \sigma T^4

Where \sigma = 5.67 x10^{-8} \frac{J}{sm^2 K^4}

If we solve for P we got:

P = A \epsilon \sigma T^4

Since we have a baseball ball considered as a sphere the superficial area is given by:

A = 4\pi r^2

Solution to the problem

And if we replace this into our equation of P we got:

P = (4\pi r^2) \epsilon \sigma T^4

And we can reorder this like that:

P = 4 \epsilon \pi \sigma r^2 T^4

We can convert the radius to meters and we got:

r= 5.2 cm*\frac{1m}{100 cm}=0.052 m

Now we can convert the temperature to Kelvin and we got:

T = 20.8 +273.15 = 293.95 K

\epsilon = 0.86 the emissivity given

And now we can replace into the formula for P and we got:

P = 4*0.86*\pi *(5.67x10^{-8} \frac{J}{s m^2 K^4}) (0.052m)^2 (293.95 K)^4

P = 12.37 \frac{J}{s} = 12.37 Watts

6 0
3 years ago
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