The solar energy strikes the deck at the rate of 1400 W on every square metre.

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

7 0

2 years ago

Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far fr

Dahasolnce [82]

Answer:

d = 0.645 m (assuming a radius of the ball bearing of 3 mm)

Explanation:

The given information is:

The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = $d_{e}$
The radius of the sun is 6.96x10⁸m = $r_{s}$

We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = $r_{b}$ .

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

$\frac{r_{s}}{r_{b}} = \frac{6.96\cdot 10^{8}m}{3\cdot 10^{-3}m} = 2.32\cdot 10^{11}$

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, $d_{s}$ :

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

7 0

2 years ago

A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas

goldfiish [28.3K]

Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400

3 0

2 years ago

The following pairs of surfaces are pushed together with the same amount of force. Between which pair of surfaces will friction

Aleks [24]

A. The friction between two pieces of sandpaper is greater than
the friction between any of the pairs of surfaces.

D. Juan should decrease the mass of his go-kart. Then any force
that pushes it forward will give it greater forward acceleration.

5 0

2 years ago

Read 2 more answers

A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo

tiny-mole [99]

The answer is left and right.

4 0

2 years ago

Read 2 more answers