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3 years ago

A ball is thrown upward in the air with an initial velocity of 40 m/s. How long does it take to

Physics

1 answer:

erma4kov [3.2K]3 years ago

8 0

Answer:

You need the definition of acceleration (a=Vf-Vi/t) and 1 equation of linear motion (deltaX = Vi×t + 1/2×a×t^2). Since you know a is constant (gravity) and you know your initial Vi to be 40 m/s and your final velocity Vf to be zero (maximum height), then you can use thhe definition of acceleration to find time.

-9.81m/s^2 = (0-40m/s)/t

t = (-40)/(-9.81) s

t = 4.077s

Now that you have time, you should know all but deltaX in the equation of linear motion.

dX = (40m/s)(4.077s) + (1/2)(-9.81m/s^2)(4.077s)^2

dX = (163.099m) — (81.549m)

dX = 81.55m

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3 years ago

An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?

hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

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Required

The current I

Step two

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3 years ago

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2 years ago

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Sergio039 [100]

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3 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago