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2 years ago

Show the relation among MA, VR and n.

Physics

1 answer:

almond37 [142]2 years ago

8 0

Answer:

good luck!!! sorry I just needed the points xoxo

Explanation:

umm yeah no sorry I tried

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What is temperature? how temperature can be measured

german

Answer:

Temperature is the kinetic energy of the particles of a substance.

Explanation:

The more kinetic energy a particle has the higher it's temperature. In the case of the atmosphere, which is what we are primarily concerned with in Meteorology, we measure this using a mercury thermometer (in certain situations we use an alcohol thermometer and of course modern times have given us things like dewcells and digital thermometers but we always go back to the mercury thermometer for accuracy).

3 0

1 year ago

In Hooke's law, Fspring=kΔx, what does the ∆x stand for

monitta

Answer:

Change in Displacement

Explanation:

delta/triangle = change

x = displacement

formula (if needed): final x - initial x

7 0

2 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

According to the question the final volume becomes twice of the initial volume.

Using ideal gas law:

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

Work done by the gas during the initial isobaric expansion:

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

we have the specific heat capacity of oxygen at constant pressure as:

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

Now change in internal energy:

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

Now heat added to the system:

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

7 0

3 years ago

PLEASE HELP! I need help with this

melisa1 [442]

At time = .003 hr

90* (3*10^-3) = .27km

.005 hr

Car A = 150 * (.005) = .75km

Car B = (90 * (.005) = .45 km

Car A = 150 * (0.008) = 1.2 km

Car B = 90 * (0.008) = 0.72 km

Car A is ahead

3 0

3 years ago

If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the in

poizon [28]

Answer:

The change in the internal energy of the gas 1,595 J

Explanation:

The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)

This is summarized in the expression:

ΔU= Q - W

where the heat absorbed and the work done by the system on the environment are considered positive.

Taking these considerations into account, in this case:

Q= 500 cal= 2,092 J (being 1 cal=4.184 J)

W=500 J

Replacing:

ΔU= 2,092 J - 500 J

ΔU= 1,592 J whose closest answer is 1,595 J

The change in the internal energy of the gas 1,595 J

6 0

3 years ago

Show the relation among MA, VR and n.​

Show the relation among MA, VR and n.