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4 years ago

Are temperature and thermal energy the same thing justify your answer

1 answer:

6 0

Temperature is the measurement of the average energy of the particles in a solid, liquid or gas and thermal energy is the total energy in a set amount of solid, liquid or gas. Therefore, the temperature and thermal energy is not the same thing. They are both about the particle theory, which is a theory that all particles of solid, liquid or gas are always in motion. But the difference between the two is that temperature is the "measurement" of the particles in a solid, liquid or gas and the thermal energy is the total energy in a set amount of solid, liquid or gas.

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What is the timeline for the law of conservation of energy

kirill [66]

In 1842, Julius Robert Mayer discovered The law of conservation of Energy. It its most compact form, it it now called The first law of Thermodynamics

Energy can neither be created nor destroyed, it can only be changed to another form of energy.

4 0

3 years ago

A conducting sphere of radius r1 = 0.21 m has a total charge of Q = 2.9 μC. A second uncharged conducting sphere of radius r2 =

wolverine [178]

Answer:

what is the anser

Explanation:

4 0

3 years ago

Use the formula from Newtons second law, namely Fnet=∆t=∆P that the Net force exerted on the object is equal to the product of t

Aleksandr-060686 [28]

Answer:

Explanation:

Fnet = Δt = ΔP

Fnet = ΔP/Δt

= mv - mu / t

= m(v - u) / t

But we know v = u + at

=> a = v - u / t

Therefore,

Fnet = ma

6 0

2 years ago

All the bulb in the series circuit are on or off together?

emmainna [20.7K]

Yes they are the power goes from one to another and if one goes off another will go off too

5 0

3 years ago

An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

3 years ago