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Margaret [11]
2 years ago
14

A building is 6m high, and it is 80m from a converging camera lens. If the camera forms an image which is 6 mm high, (a) What is

the magnification? (b) How far must the camera film be behind the lens for the image to be formed?
Physics
1 answer:
Lostsunrise [7]2 years ago
5 0

Answer:

Explanation:

Two aircrft pand qare flyingatthe same speed 300m the diraction along tothe diraction of the veloctiy

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True or false questions for work and power.​
dalvyx [7]

Answer:

lo escribes en español?

Explanation:

no hablo ingles

5 0
2 years ago
Which of the following statements best describes the second law of
Serga [27]

Answer:

B

Explanation:

Heat flows from hot to cold to lower the temperature of hot areas and increase temperature of cold areas. The end result is that the 2 areas have the same temperature, thus increasing entropy.

7 0
3 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
The graph represents velocity over time...
Nady [450]

Answer:

where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please

8 0
2 years ago
How do the lungs and heart transport oxygen in the cardio respiratory system
harina [27]

The lungs hold air that is taken in. Oxygen gas noticeable all around moves into the blood. The heart pumps to transports this oxygenated blood to cells in the body that need it to deliver vitality.

7 0
3 years ago
Read 2 more answers
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