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3 years ago

A complete path through which charge can flow is an electric what?

Physics

2 answers:

grigory [225]3 years ago

5 0

<span>A complete path through which charge can flow is an "Electric Circuit"

Hope this helps!</span>

kvasek [131]3 years ago

3 0

Answer: The correct answer is "electric circuit"

Explanation:

Electric circuit: It is a path through which the charges can flow easily. It is a path for transmitting electric current.

It is a continuous path through which the electrons can flow in the closed circuit. The flow of charges constitutes the current in the circuit. The battery makes the charges flow in the circuit or voltage source.

If the current is flowing in the electric circuit then it is said to be closed electric circuit.

If the current is not flowing in the electric circuit then it is said to be open electric circuit.

Therefore, a complete path through which charge can flow is an electric circuit.

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An astronaut of mass m in a spacecraft experiences a gravitational force F=mg when stationary on the launchpad.

zaharov [31]

gravitational force is the attraction force of earth on an object which is near the surface of earth

It will not depend on the velocity or acceleration of earth

So it will not change while an object is stationary or it is moving with some acceleration

So here we will say that force will remain the same

$F = mg$

7 0

4 years ago

Two forces that make up 3rd law of motion can: 1)Act on the same object 2)do not act on same time 3)Act on different objects 4)D

Ede4ka [16]

Answer:

3) Act on different objects

Explanation:

Newton's third law of motion states that for every action there is an equal and opposite reaction.

Action-reaction force pairs make it possible for fishes to swim, birds to fly, cars to move, etc.

Basically, action-reaction force pairs are characterized by the following statements;

I. The forces are the same type such as magnetic force, contact force or gravitational force.

II. The forces are equal in magnitude but opposite in direction.

III. They act up on two different bodies.

Hence, the two forces that make up 3rd law of motion can act on different objects.

For example, while driving down the road, a bug strikes the windshield of a vehicle (Action) and makes a quite obvious mess in front of the face of the driver of the vehicle (Reaction) i.e the bug hit the vehicle and the vehicle hits the bug.

5 0

3 years ago

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

4 years ago

Which of the following correctly describes the first animals that appeared?

Dvinal [7]

A these were large creatures similar to dinosaurs

7 0

4 years ago

A 42.5 g piece of aluminum (which has a molar heat capacity of 24.03 j/ocmol) is heated to 82.4oc and dropped into a calorimeter

Leno4ka [110]

1. Use the equation q = nC∆T
n = mols of aluminum = 42.5g/ 6.98g/mol
C = molar heat capacity = 24.03 j/Cmol
∆T = change in T from what it was before placed in calorimeter to after =
24.9C-82.4C (because water final and metal will be same temp.)
plug in to calculate q of metal =
q = (42.5/6.98)*(24.03J)(24.9-82.4)
q = -2176.5498 J
qmetal = -q of water
plug in values for water
-(-2176.5498 J) = mC∆T (m = mass of water in grams)
m = q/C∆T
∆T=24.9-22.3 C = 2.6
2176.5498 J/(2.6x4.184) = m = 200.2714 g

4 0

3 years ago

Read 2 more answers