2 years ago

What is the difference between convex and concave mirrors and lens?

Physics

1 answer:

Sidana [21]2 years ago

5 0

Concave makes things smaller and convex makes things bigger

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A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a

valkas [14]

The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
$\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}$

The rate of change of the radius is given as
$\frac{dr}{dt} =-2 \, \frac{ft}{s}$
Therefore
$\frac{dA}{dt} =-4 \pi r \, \frac{ft^{2}}{s}$

When r = 10 ft, obtain
$\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi \, \frac{ft^{2}}{s}$

Answer: - 40π ft²/s (or - 127.5 ft²/s)

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3 years ago

The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat

Nonamiya [84]

Answer:

E=0

Explanation:

The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero

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3 years ago

What kind of tectonic plate boundary must have existed in the past where copper deposits are found, based on the way they are fo

Maksim231197 [3]

Answer:

older plates that are subducting almost perpendicularly, and when the location is far away from the edge of other tectonic plates, are the most likely areas for copper deposits to form

Explanation:

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3 years ago

What temperature does water freeze at in celsius and fahrenheit?

andre [41]

32 degrees Fahrenheit
And
0 degrees Celsius

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3 years ago

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Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima

Svet_ta [14]

Answer:

$5.241\ \text{m/s}$

Explanation:

m = Masa del coche

g = Aceleración debida a la gravedad = $9.81\ \text{m/s}^2$

h = Altura = $1.4\ \text{m}$

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}$

La velocidad máxima que puede alcanzar el coche es $5.241\ \text{m/s}$ .

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3 years ago