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iogann1982 [59]

4 years ago

11

The intensity of sunlight incident on the surface of a solar panel on your roof is 1000 W/m^2. Calculate the sunlight energy abs

orbed by the panel in one hour if the length of the panel is 6.00 m and its width is 4.00 m and the sunlight is incident perpendicularly on the panel.

1 answer:

melisa1 [442]4 years ago

5 0

Answer:

8.64 x 10^7 J

Explanation:

Intensity of sunlight, I = 1000 W/m^2

Length of panel, L = 6 m

Width of panel, W = 4 m

Area of the panel, A = L x W = 6 x 4 = 24 m^2

time, t = 1 hour = 3600 second

Energy = Intensity x area of panel x time

E = 1000 x 24 x 3600 = 8.64 x 10^7 J

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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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4 years ago

A space station worker found herself floating free 100 meters from the space station because her safety line became unhooked

yuradex [85]

Answer:

Explanation: is that suppose to be the question or are you just saying?

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3 years ago

Direction of current in a solenoid depends on the direction of magnetic field

cestrela7 [59]

Explanation:

Its direction depends on the direction of the current. ... When the current flows through the solenoid in the clockwise direction, then the magnetic lines of force inside or center of the coil will be along the axis inwards according to Fleming's right hand rule.

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3 years ago

a. Define Valency. b. Write the symbol and valency of the following : (i) Oxygen (ii) potassium 7. a. Magnesium burns in Oxygen

sweet-ann [11.9K]

Answer:

Oxygen-Symbol-O Potassium Symbol-K

6 0

4 years ago

Read 2 more answers

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

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3 years ago