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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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Calculate the total number of Cl atoms in 150mL of liquid Ccl4 (d=1.589g/mL)<br>​
GalinKa [24]

Answer:

The total number of Cl atoms in 150mL of liquid CCl4 is 3.73*10²⁴.

Explanation:

First you must determine the mass of CCL4 present in 150mL of CCl4. Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance, whose expression for its calculation is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

In this case, the density value of d = 1.589 g/mL. Then, being the volume equal to 150 mL, the value of the mass can be calculated as:

mass= density*volume

mass=1.589 g/mL * 150 mL

mass= 238.35 g

Now, being the molar mass of CCl4 154 g/mol, the number of moles that 238.35 g represents is calculated as:

moles=\frac{238.35 g}{154 \frac{g}{mol} }

moles= 1.55

1 mole of the compound CCl4 contains 4 moles of Cl. Then, using a simple rule of three, it is possible to calculate the number of moles of Cl that 1.55 moles of CCl4 contain:

moles of Cl=\frac{1.55 moles of CCl_{4} *4 moles of Cl}{1 mole of  CCl_{4} }

moles of Cl= 6.2

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance.  In this case it can be applied as follows: if 1 mole of Cl contains 6.023*10²³ atoms, 6.2 moles of Cl how many atoms does it contain?

atoms of Cl=\frac{6.2 moles*6.023*10^{23} atoms}{1 mole}

atoms of Cl= 3.73*10²⁴

<u><em>The total number of Cl atoms in 150mL of liquid CCl4 is 3.73*10²⁴.</em></u>

8 0
2 years ago
What is the purpose of the overrunning clutch in the starter drive?
nadya68 [22]
<span>"prevent the engine from over speeding the armature"
hopes this help :) :D :)</span>
5 0
3 years ago
Charged beads are placed at the corners of a square in the various configurations shown in
Akimi4 [234]

Answer:

The consecutive charge configuration has a more intense field than alternating

Explanation:

In each corner we place a different account there are only two different settings, see attached.

In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field

In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is

          E = 2 k q / ra2 a cos 45

The consecutive charge configuration has a more intense field than alternating

8 0
3 years ago
Give an example of a force applied to an object that does not change the object's velocity. Why does the object's velocity not c
Xelga [282]

Answer:

A chair at rest on the floor has two forces acting on it its own weight that pulls it downward and the floor pushing upward on the chair, both of these forces are acting on it but the net force is 0, so the chair remains at rest and its velocity stays at 0.

8 0
1 year ago
Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha
tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

         R_cm = 1 / M   ∑  x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

       R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

      M = m_earth + m_moon

     

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

      M = 6.0535 10₂⁴24 kg

we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

      R_cm = 4.66 10⁶ m

4 0
2 years ago
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