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3 years ago

How to tell if two objects are in thermal equilibrium

Physics

1 answer:

QveST [7]3 years ago

6 0

1.) the objects have the same temperature

this is the answer because thermal equilibrium is when two substances touch and the temperature becomes the same.

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How to convert sin angles with values of 90 degrees, 60, 45, 30, 0 to cos degrees.

Inga [223]

Answer:

ghjfglghhukfukxhhfjf

6 0

3 years ago

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

3. Liquid nitrogen at 90 K, 400 kPa flows into a probe used in a cryogenic survey. In the return line the nitrogen is then at 16

devlian [24]

Answer:

Specific heat transfer = 236.16 kJ/kg

Ratio of return velocity to inlet velocity = 0.80

Explanation:

Given

Temperature of liquid nitrogen, T1 = 90 K

Pressure of liquid nitrogen, P1 = 400 kPa

Temperature of nitrogen, T2 = 160 K

Pressure of nitrogen, T2 = 400 kPa

A(e) = 100 A(i)

To solve, we use the formula

h(i)+ 1/2v(i)² + q = h(e) + 1/2v(e)² + q

The mass flow is

m = m(i) = m(e)

m = (Av/V)i = (Av/V)e

Ratio of return velocity to inlet velocity is

v(e) / v(i) = A(i)/A(e) * V(e)/V(i)

v(e) / v(i) = 1/100 * V(e)/V(i)

From the saturated Nitrogen table, at 100 K, we have

h(i) = h(f) = -73.2

v(i) = v(f) = 0.001452

From the saturated Nitrogen table again, at 160 K and 400 kPa

h(e) = 162.96 kJ/kg

v(e) = 0.11647 m³/kg

Substituting these in the formula, we have

v(e) / v(i) = 1/100 * 0.11647/0.001452

v(e) / v(i) = 1/100 * 80.2

v(e) / v(i) = 0.80

Energy equation is given by

q + h(i) = h(e)

q = h(e) - h(i)

Now, calculating specific heat transfer

q = 162.96 - -73.2

q = 236.16 kJ/kg

6 0

3 years ago

Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardr

Margaret [11]

Answer:

a. d₁/d₂ = 1.09 b. 0.054 mW

Explanation:

a. What is the ratio of the diameter of the first student's eardrum to that of the second student?

We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So, I ∝ I/d²

I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.

Given that I₂ = 1.18I₁

I₂/I₁ = 1.18

Since I₁/I₂ = d₂²/d₁²

√(I₁/I₂) = d₂/d₁

d₁/d₂ = √(I₂/I₁)

d₁/d₂ = √1.18

d₁/d₂ = 1.09

So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09

b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

We know intensity, I = P/A where P = acoustic power and A = area = πd²/4

Now, P = IA

= I₂A₂

= I₂πd₂²/4

= 1.18I₁πd₂²/4

Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

So, P = 1.18I₁πd₂²/4

= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

= 0.691244π × 10⁻⁴ W/4 =

2.172 × 10⁻⁴ W/4

= 0.543 × 10⁻⁴ W

= 0.0543 × 10⁻³ W

= 0.0543 mW

≅ 0.054 mW

3 0

3 years ago

In what way were augusts<br> comte and immanual Kant<br> alike

vodka [1.7K]

1. they were both educated in france
2. they both felt that the end of the monarchy was inevitable
3. they were both artist
4. they both felt that science would improve the world

5 0

3 years ago

Read 2 more answers