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scZoUnD [109]
3 years ago
11

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

velocity vectors of the plane at a later time are given by r=(1.21x103i+3.45x104;)m and v= (2 i-3.5j) m/s The magnitude, in meters, of the plane's displacement from the origin is:_________
a. 2.50 x104
b. 1.45 x 104
c. 3.45x104
d. 2.5x103
e. none of the above
Physics
1 answer:
Contact [7]3 years ago
6 0

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m

hence, the displacement of the airplane is 3.45*10^4 m

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A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
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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
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