How do you know there is current in a circuit?

Which of the following surfaces reflects the most light?

Lunna [17]

Aluminum foil reflects more light

3 0

3 years ago

Read 2 more answers

4 points

zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

$I = k\frac{1}{r^2}$

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

$I_2 = k\frac{1}{(2r)^2}$

Dividing both the equation we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4} = 25lx$

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

$I_2 = k\frac{1}{(3r)^2}$

Dividing equation 1 and 3 we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9} = 11.11lx$

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0

3 years ago

Calculate the force of gravity on a 1–kilogram box located at a point 1.3 × 107 meters from the center of Earth if the force on

Sati [7]

Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.

Look at the formula for the gravitational force:

F = G m₁ m₂ / R² .

If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:

(6.1 N) / (2.5 kg) = (F) / (1 kg)

Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)

Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .

5 0

3 years ago

Mercury is in the 80th position in the periodic table. How many protons does it have?

Verdich [7]

Mercury has 80 protons. Ironic?

7 0

2 years ago

Read 2 more answers

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$