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2 years ago

What are applications of zeroth law of thermodynamics?

Physics

1 answer:

Harrizon [31]2 years ago

6 0

Answer:

Applications of zeroth law of thermodynamics:

1. When we get very hot food, we wait to make it normal. In this case, hot food exchanges heat with surrounding and brings equilibrium.

2. We keep things in the fridge and those things come equilibrium with fridge temperature.

3. Temperature measurement with a thermometer or another device.

4. In the HVAC system, sensors or thermostats are used to indicate temperature. It always comes in a thermal equilibrium with room temperature.

5. If you and the swimming pool you’re in are at the same temperature, no heat is flowing from you to it or from it to you (although the possibility is there). You’re in thermal equilibrium.

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A submarine has a "crush depth" (that is, the depth at which

sergey [27]

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

$P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm$

So, the approximate pressure is equal to 40.69 atm.

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2 years ago

Is it true that acceleration occurs when there is a change in speed

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3 years ago

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

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3 years ago

Explain the correspondence that lets us easily translate between linear motion and rotational motion. What are the linear analog

RideAnS [48]

Explanation:

The linear analog of angle is angle itself.

The linear analog of angular velocity is linear velocity.

ω is angular velocity, therefore linear velocity is given by v

∴ for linear velocity, $v^{2} = u^{2}+2.a.S$

for angular velocity, $\omega_{f}^{2} = \omega _{i}^{2}+2.a.S$

The linear analog of angular acceleration is acceleration.

α is angular acceleration whereas as a is linear acceleration.

∴ for linear acceleration, v = u + a.t

for angular acceleration, $\omega_{f}= \omega _{i}+\alpha .t$

The linear analog of moment of inertia is mass.

I is moment of inertia and m is mass,

∴ for linear analog, F = m.a

for angular analog, τ - I.α

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3 years ago

What are applications of zeroth law of thermodynamics?​

What are applications of zeroth law of thermodynamics?