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3 years ago

Write 3 major differences between a shadow and a image?

Physics

2 answers:

Andru [333]3 years ago

4 0

Answer:

We know that an image is a picture of an object while shadow is formed an object comes in between the surface and light. ... The formation of image takes place when the light rays are reflected by an object. The formation of shadow takes place when the light falls on an opaque object.

Phantasy [73]3 years ago

3 0

Explanation:

Image Shadow

2. We are able to see images when light refraction enters our eyes.

In the case of shadow no light enters our eyes.

3. Image contains colour, structure etc. of the object.

Shadow does not give any information about the object as it is colourless

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The Global Positioning System (GPS) is a constellation of about 24 artificial satellites. The GPS satellites are uniformly distr

Liono4ka [1.6K]

Answer:

b) 3.72m/s²

c) 9.33*10^5

d) 9.33*10^5

e) 11.85 hrs

Explanation:

a) to confirm that gEarth is about 98 m/s².

Let's use the formula:

$gEarth= \frac{G*M}{R^2}$

$= \frac{6.67*10^-^1^1*5.972*10^2^4}{(6378*10^3)^2}$

= 9.78 m/s²

=> 9.8m/s²

b) Given:

$m = 6.417*10^2^3$

r = 2106 miles

$g_Mars = \frac{G*M}{R^2}$

$= \frac{6.67*10^-^1^1*6.417*10^2^3}{(2106*1.61*10^3)^2}$

=3.72 m/s²

c) we use:

$F = \frac{G*M*m}{R^2}$

$=\frac{6.67*10^-^1^1*5.972*10^2^4*1630*10^3}{((20000+6378)*10^3)^2}$

$= 9.33*10^5 N$

d) Let's take the force of gravitybon earth due to satellite as our answer in (c) because the Earth's gravitational force on a GPS satellite and the force of gravity on a GPS satellite on earth are equal and opposite (two mutual forces).

$F = 9.33*10^5 N$

e) In a circular motion,

Gravitional force = Centripetal force.

$\frac{GM*m}{R^2}=\frac{m*v^2}{R}$

$\frac{GM}{R}= v^2$

Solving for v, we have

$v= \sqrt{\frac{6*67*10^-^1^1*5.972*10^2^4}{(20000+6278)*10^3}}$

v = 3886m/s

Therefore,

v = 2πR/T

$3886 = \frac{2*pi*(20000+6378)*10^3}{T}$

Solving for T, we have:

T = 42650seconds

Convert T to hours

T = 42650/60*60

T = 11.86hrs

6 0

4 years ago

A 12 kg block is released from the top of an incline that is 5.0 m long and makes an angle of 40.0º to the horizontal. A force o

Allushta [10]

Answer:

do u have a photo that comes w/ this? so i could help more :) ?

Explanation:

8 0

3 years ago

Please help u guys acellus sucks

34kurt

Answer:

6.0cm

Explanation:

Given

focal length = 15.0cm

object distance = 10.0cm

Required

Image distance v

Using the formula

1/f = 1/u + 1/v

1/15 = 1/10+1/v

1/v = 1/15 + 1/10

1/v = 2+3/30

1/v = 5/30

v = 30/5

v = 6.0cm

Hence the image distance is 6.0cm

7 0

3 years ago

Is dimensionally correct equation must be physically correct . Explain your logic behind physically correct equation ?

anyanavicka [17]

An equation can be dimensionally correct but that does not mean that it is physically correct. Explanation is given in the photo.

8 0

1 year ago

what distance is a book from the floor if the book contains 196 joules f potential energy and has a mass of 5 kg?

AleksAgata [21]

E=mgh. 196=5kg*9.81m/s^2*h. So h=196/(5*9.81)=4m

5 0

4 years ago