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3 years ago

9

What allows the bimetallic coil to turn on or off a heating or cooling system? The two metals contract the same amount. The ther

mostat is set at a particular temperature. The two metals expand differently. The coil comes in contact with air.

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Answer:

The two metals expand differently.

Explanation:

The bimetallic strip has two metal strips positioned like a bridge, these strips connect the electrical circuit to the heating system. When these strips are linear or "down" they allow the electricity to move through the circuit to the heating system to turn the heat on. When the strips are "up" the disconnect the electricity flow, thus turning the heating system off, thus the room becomes cool/cold.

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A glider of mass 5.0 kg hits the end of a horizontal rail and bounces off with the same speed, in the opposite direction. The co

NeTakaya

To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.

Let's start by defining acceleration based on speed and time, that is

$a = \frac{v}{t}$

On the other hand according to Newton's second law we have to

F=ma

where

m= Mass

a = Acceleration

Replacing the value of acceleration in this equation we have

$F=m(\frac{v}{t})$

Substituting with our values we have

$100N=(5Kg)\frac{v}{0.2}$

Re-arrange to find v

$v=\frac{100*0.2}{5}$

$v = 2m/s$

Therefore the speed of the glider is 2m/s

5 0

3 years ago

Runner 1 has a velocity of 10 m/s west. Runner 2 has a velocity of 7 m/s east. From the frame of reference of runner 2, what is

Vsevolod [243]

Answer:

<em>17 m/s west</em>

Explanation:

Runner 1 has velocity = 10 m/s west

runner 2 has velocity = 7 m/s east

From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of

velocity = 10 m/s + 7 m/s = <em>17 m/s west</em>

5 0

3 years ago

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi

zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} = mass flow rate\\\rho = Density\\V = Velocity$

values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3V_1 = 40m/s$

$A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s$

PART A) Applying the flow equation

$\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2$

7 0

3 years ago

A blue car with mass mc = 427 kg is moving east with a speed of vc = 20 m/s and collides with a purple truck with mass mt = 1282

polet [3.4K]

Answer:8540 kg-g/s

Explanation:

Given

mass of blue car $m_c=427 kg$

velocity of blue car $v_c=20 m/s$

mass of the truck $m_t=1282 kg$

speed of truck $v_t=13 m/s$

After collision they stick and lock together

Let v be the velocity of combined system at angle \theta from vertical

Conserving momentum in east direction

$m_c\times v_c=(m_c+m_t)v\cos \theta$

$427\times 20 =1709\times v\cos \theta$ ------1

Conserving Momentum in Y direction

$m_t\times v_t=(m_c+m_t)v\sin \theta$

$1282\times 13 =1709\times v\sin \theta$ -------2

squaring and then adding 1 & 2 we get

$(8540)^2+(16666)^2=(1709)^2\cdot v^2$

v=10.95 m/s

initial momentum of car $=427\times 20=8540 kg-m/s$

6 0

4 years ago

One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist

Svetach [21]

<h3>Answer:</h3>

(a) <u>i₁ = 0.03818 A = 38.18 mA</u>

(b) downward

(c) <u>i₂ = 0.01091 A = 10.91 mA</u>

(d) rightward

(e) <u>i₃ = 0.02727 A = 27.27 mA</u>

(f) leftward

(g) <u>Eₐ = 3.818 Volts</u>

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0

E₂ - i₃R₃ - i₁R₁ = 0

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

<u>i₃ = 0.02727 A</u>

using this value in eqn of i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

<u>i₂ = 0.01091 A</u>

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

<u>i₁ = 0.03818 A</u>

<u></u>

(a)

<u>i₁ = 0.03818 A = 38.18 mA</u>

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>downward</u>

(c)

<u>i₂ = 0.01091 A = 10.91 mA</u>

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>rightward</u>

(e)

<u>i₃ = 0.02727 A = 27.27 mA</u>

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>leftward</u>

<u>(g)</u>

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

<u>Eₐ = 3.818 Volts</u>

3 0

4 years ago