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3 years ago

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How efficient are the small and large scale solar-power systems used in individual homes and industrial settings? What is the en

vironmental impact of the generation of solar power?
In detail plz

1 answer:

Leviafan [203]3 years ago

4 0

Answer:

$\color{Blue}\huge\boxed{Answer}$

<em>The potential environmental impacts associated with solar power—land use and habitat loss, water use, and the use of hazardous materials in manufacturing—can vary greatly depending on the technology, which includes two broad categories: photovoltaic (PV) solar cells or concentrating solar thermal plants (CSP).</em>

Explanation:

I just answer the second question

You might be interested in

How much potential energy does a 50-N box have when lifted at a height of 1.5M?

nikitadnepr [17]

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (<em>mg</em>) of the box is given as the unit is <em>Newton</em>, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)

In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5 = <em>75 J (Option A)</em>

3 0

4 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

The bowl now contains . Milliters of oil

Sergeeva-Olga [200]

Data????????????????????????????

7 0

3 years ago

Read 2 more answers

A car slows down from 30 m/s to rest in a distance of 85 m. What was its acceleration?

ziro4ka [17]

120 acceleration oh yeah i am sure about this

4 0

4 years ago

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart.

serious [3.7K]

Answer:

$1.6\times 10^{-7}$ N

$2.4\times 10^{-7}$ N

Explanation:

$i_{1}$ = 1 A

$i_{2}$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}$

$F = (10^{-7})\frac{2(1)(4)}{5}$

$F = 1.6\times 10^{-7}$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))$

$B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))$

$B = 2.4\times 10^{-7}$

5 0

3 years ago