All categories

3 years ago

A(n) ____________ image cannot be projected and forms where light rays appear to originate.

Physics

2 answers:

IrinaVladis [17]3 years ago

6 0

A virtual image is the answer.

Sedbober [7]3 years ago

6 0

<h3><u>Answer;</u></h3>

Virtual image

A<em><u> virtual image</u></em> cannot be projected and forms where light rays appear to originate.

<h3><u>Explanation</u>;</h3>

<em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
<em><u>A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.</u></em>

You might be interested in

The fall of a body on the earth surface cannot be a complete free fall ? why<br>

anzhelika [568]

It can never be in a total free fall due to air resistance and terminal velocity due to mass

8 0

2 years ago

Calculate the momentum of a 1,500 kg car traveling at 6 m/s.

mamaluj [8]

Answer: 9000 kgm/s

Explanation:

Mass of car = 1500 kg

Speed by which car moves = 6 m/s. Momentum of the car = ?

Recall that:

Linear momentum = Mass x Speed

= 1500kg x 6m/s

= 9000 kgm/s

Thus, the linear momentum of the car is 9000 kgm/s

3 0

3 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$