Ask question

Ask question

All categories

3 years ago

8

Using a dish-shaped mirror, a solar cooker concentrates the sun's energy onto a pot for cooking. A cooker with a 1.2-m-diameter

dish focuses the sun's energy onto a pot with a diameter of 25 cm What is the intensity at the base of the pot

1 answer:

Alekssandra [29.7K]3 years ago

3 0

Answer:

26500 W/m²

Explanation:

Given that:

The diameter of the dish d = 1.2 m

Assuming the Solar power P capture by the dish = 1300 W

The surface area of the pot is calculated by using the formula:

$= \pi r^2$

$\pi \dfrac{d^2}{4}$

where; the diameter = 25 cm

Area = $\pi \dfrac{d^2}{4}$

Area = $\pi \dfrac{0.25^2}{4}$

Area = 0.049 m²

The intensity is calculated by using the formula

I = power/area

I = 1300/0.049

I = 26530.6 W/m²

I ≅ 26500 W/m²

You might be interested in

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

3 years ago

What are the basic si units for the speed of light?.

DanielleElmas [232]

C=meters/second or C=m/s

7 0

2 years ago

What is the scientific definition of tracked?

dexar [7]

To find or discover by investigation?

8 0

3 years ago

To accelerate, an object must *

Sonja [21]

Answer: Not 100% sure but I think it’s C.

Hope this helps! ^^

4 0

3 years ago