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3 years ago

Explain the relationship of charles law to absolute zero

Physics

1 answer:

vivado [14]3 years ago

8 0

Chalres law and absolute zero . in 1780s, by experiments

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A locomotive is pulling 15 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (w

VMariaS [17]

Answer:

298,220 N

Explanation:

Let the force on car three is T_23-T_34

Since net force= ma

from newton's second law we have

T_23-T_34 = ma

therefore,

T_23-T_34 = 37000×0.62

T_23= 22940+T_34

now, we need to calculate

T_34

Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2

F= ma

So, F= 12×37000×0.62= 22940×12= 275280 N

T_23 =22940+T_34= 22940+ 275280= 298,220 N

therefore, the tension in the coupling between the second and third cars

= 298,220 N

3 0

3 years ago

The drawing shows two thermometers, A, and B, whose temperatures are measured in °A and °B. The ice and boiling points of water

Rama09 [41]

Unf there's no diagram. but this looks like a sort of celsius to fahrenheit temp scale conversion sort of problem.

6 0

3 years ago

if a free falling rock were equipped with a speedometer, by how much would it's speed readings increase with each second IF it w

Kaylis [27]

Accelleration(a) is changing of velocity in second.
free fall a = g
speed increase = a = g = 20 (m/s) / s

7 0

3 years ago

Which of these is a landform created due to erosion by glaciers? Delta Horn Sandbars Spit

Marina CMI [18]

Answer:

The answer is option B, horn.

Explanation:

The landform horn was formed because when glaciers erode down to form a sharp peak which is called a horn. All of the other options are false because none of those land forms are formed because of the erosion of glaciers

4 0

3 years ago

A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected

vekshin1

Answer:

(A) $9.14\times 10^{-9}sec$

(B) $6.20\times 10^{-3}A$

Explanation:

We have given inductance $L=5.41\mu H=5.41\times 10^{-6}H$

Resistance $R=0.949kohm=0.949\times 10^3ohm$

Time constant of RL circuit is equal to $\tau =\frac{L}{R}$

$\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec$

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

$i=i_0(1-e^{\frac{-t}{\tau }})$

According to question

$0.799i_0=i_0(1-e^{\frac{-t}{\tau }})$

$e^{\frac{-t}{\tau }}=0.201$

${\frac{-t}{\tau }}=ln0.201$

${\frac{-t}{5.7\times 10^{-9} }}=-1.6044$

$t=9.14\times 10^{-9}sec$

(b) Current at $t=\tau sec$

$i=i_0(1-e^{\frac{-t}{\tau }})$

$i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})$

$i=6.20\times 10^{-3}A$

3 0

3 years ago