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2 years ago

Hello people ~

Physics

1 answer:

Fantom [35]2 years ago

3 0

Explanation:

In di electric medium

The force is

$\sf F_{dE}=\dfrac{1}{4K\pi\epsilon}\dfrac{q_1q_2}{R^2}$

So the force on air is decreased k times

Option C is correct

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Which of the following is not a characteristic of electromagnetic light waves?

SOVA2 [1]

Answer:

they cant travel through a vacuum

7 0

2 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

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points

Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

5 0

3 years ago

Saul convert 2.392 hectoliters to liters what should his new number be?

murzikaleks [220]

His new number should be: 239.2 Liters

4 0

3 years ago

Imagine a Rip-van-Winkle type who lives in the mountains. Just before going to sleep he yells "WAKE UP" and the sound echoes off

stiv31 [10]

Answer:

4939200 m

Explanation:

v = Velocity of sound in air = 343 m/s (general value)

For the echo to reach Rip van Winkle it must cover the distance to the mountain twice. So, time taken for the sound traveling to the mountain once

$t=\dfrac{8}{2}=4\ hr$

Distance is given by

$s=vt\\\Rightarrow s=343\times 4\times 60\times 60\\\Rightarrow s=4939200\ m=4939.2\ km$

The distance to the mountain is 4939200 m

6 0

3 years ago