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iVinArrow [24]
3 years ago
11

How does a real image differ from a virtual image?

Physics
2 answers:
Snezhnost [94]3 years ago
6 0

Answer:

one contains datta that can take up hard drive or disc space weras a picture takes up space on walls or photo albums

Explanation:

goldfiish [28.3K]3 years ago
3 0
Real image is formed when the rays of light after reflection or refraction actually meet at some point whereas a virtual image is formed when the rays of light after reflection or refraction appear to meet at a point
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What’s 4.5 inches into centimeters?
marshall27 [118]

Answer:

11.43 cm

Explanation:

yes

3 0
4 years ago
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What is the difference between temperature and heat<br>​
uysha [10]

Explanation:

Heat is a form of energy that can transfer from hot body to cold body.Temperature is the degree of hotness and coldness of a body.

Heat is measured in joules, but temp. is measured in Kelvin

7 0
3 years ago
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An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +32.6°. When the potentia
Reil [10]

Answer:

V=-8.35v

Explanation:

Pls see attached file

6 0
3 years ago
A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0
3 years ago
A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif
Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

R=\frac{v^{2}sin(2\theta)}{g}

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

gR=v^{2}sin(2\theta)

\frac{gR}{v^{2}}=sin(2\theta)

sin^{-1}(\frac{gR}{v^{2}})=2\theta

\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}

so now we can substitute the given data:

\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

tan \theta = \frac{h}{45.5m}

so we can solve this for h, so we get:

h=45.5m*tan(0.05614^{o})

which yields:

h=0.0445m

or

h=4.45cm

5 0
3 years ago
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