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3 years ago

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Consider a metal rod of uniform temperature. Is it possible that heat spontaneously flow from one end of the rod to the other en

d so that they end up in different temperatures? Why?

1 answer:

raketka [301]3 years ago

4 0

Answer:

No its not possible in that yes heat flow is due to temperature different but with time equilibrium in temperature is reached hence there is no temperature difference at both ends at this time

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Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt

Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

2d. $\rho gh$ again, what a surprise! $800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa$

3. Yet again, $\rho gh$ . $1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa$

4 0

2 years ago

PLEASE HELP WILL GIVE 10 POINTS!!!!!

Evgen [1.6K]

Its readily available
No Harmful emissions
Environment friendly
Renewable

Hydrogen isn't very good fuel source due to its high flammability and can create a nasty mini hydrogen bomb. <span />

8 0

3 years ago

Read 2 more answers

If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated

sp2606 [1]

Answer:

$611.8$ nm

Explanation:

$n_{oil}$ = Index of refraction of soap bubble = 1.33

$t$ = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

$\lambda$ = wavelength of light = ?

$m$ = order = 0

For reflection , the necessary condition is

$2 n_{oil} t = (m + 0.5) \lambda$

$2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda$

$\lambda = 6.118\times 10^{-7}$

$\lambda = 611.8$ nm

7 0

3 years ago

The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid

IgorLugansk [536]

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x $x velocity^{2}$

incoming kinetic energy = 0.5 x m x $x u^{2}$

coefficient of restitution = $\frac{v}{u}$

0.72 = $\frac{v}{u}$

v = 0.72u

therefore the outgoing kinetic energy = 0.5 x m x $(0.72u)^{2}$

outgoing kinetic energy = 0.5 x m x $0.5184 x u^{2}$

outgoing kinetic energy = 0.5184 (0.5 x m x $x u^{2}$ )

recall that 0.5 x m x $x u^{2}$ is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

5 0

3 years ago

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solu

Anastasy [175]

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

7 0

3 years ago