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3 years ago

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An object is 3.0 cm from a concave mirror, with a focal length of 1.5 cm. Calculate the image distance. Remember to include your

data, equation, and work when solving this problem.

1 answer:

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

Construct a quadrilateral ABCD, where

Construct a quadrilateral ABCD, whereAB = 4 cm, BC = 5 cm, CD = 6.5 cm and angle B = 105° and angle C = 80°

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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

3 years ago

Which is a disadvantage of using wind as an energy source?

lesantik [10]

Answer:

it produces

Explanation:

a lot of waste

5 0

2 years ago

Read 2 more answers

Please help me .finish this paper

Evgesh-ka [11]

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>-</u><u>1</u><u>:</u><u>-</u>

$\boxed{\sf \dfrac{10\times 1000}{60\times 60}}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>2</u>

$\boxed{\sf Sodium\:and\:Potassium}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>3</u>

$\boxed{\sf 320m}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>4</u>

$\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>5</u>

$\boxed{\sf Mg_3N_2}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>6</u>

$\boxed{\sf Grapes\:and\:Rambutan}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>7</u>

$\boxed{\sf {}^{}_{}N}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>8</u>

$\boxed{\sf Galactuse}$

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>9</u>

$\boxed{\sf Y-X}$

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7 0

3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

<em>T</em> - 25 N = (8 kg) (2.4 m/s²)

<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N

5 0

2 years ago