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VashaNatasha [74]
3 years ago
8

Help ;-;

Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Answer:

all qn 1,2,3 have same answer ,. Yes,. hope it helps

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melisa1 [442]

Answer:

heat is the transfer of thermal energy from a system to its surroundings or from ... It is very important to know that, in science, heat and temperature are not the same thing. ... Have you noticed that when you put a cold, metal teaspoon into your hot cup of ... AIM: To investigate which materials are the best conductors of heat.

Explanation:

5 0
3 years ago
How to do problems 7-10
Stella [2.4K]
I cant see it very clear
8 0
3 years ago
1500 kg Peugeot car is traveling at 16.67 m s ⁻¹ and accelerates to 30.56 m s ⁻¹ for 2 minutes. Calculate the impulse of the car
Elis [28]

Answer:

2084 kg*m/s

Explanation:

Impulse is change in momentum

Mathematically;

Impulse, J= F*t=mΔv   where

F= ma = 1500 * { 30.56 - 16.67}/2*60

F= 1500 *0.11575

F=  174 N

J=F*t

J= 174*120*0.1

J= 2084 kg*m/s

5 0
3 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
3 years ago
40 POINTS!!!
Feliz [49]

1.) B.

2.) A.

3.) A.

If these are wrong please delete.

Thank You!

<3

4 0
3 years ago
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