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Vladimir79 [104]
3 years ago
9

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

a constant velocity of 10 m/s in the positive x-direction. At t=3 s, the center of mass of the two objects is at the position x=
Physics
1 answer:
love history [14]3 years ago
4 0

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em>  after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}

x_{cm} = \dfrac{8  \times0 + 16 \times  30}{8 + 16} = 20

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

brainly.com/question/18557256

brainly.com/question/20714030

brainly.com/question/17088562

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An object is placed near a concave mirror having a radius of curvature of magnitude 60 cm. How far should you place the object f
Viefleur [7K]

Answer:

u = 18 cm

Explanation:

given,

radius of curvature = 60 cm

magnification of mirror = 2.5

distance of object  = ?

R = 2 f

f = R/2

f = 60/2 = 30 cm

m = -\dfrac{v}{u}

2.5 = -\dfrac{v}{u}

v = -2.5 u

now,

Using mirror formula

\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}

\dfrac{1}{30} = \dfrac{1}{-2.5u} + \dfrac{1}{u}

\dfrac{1}{30} = \dfrac{0.6}{u}

u = 0.6 x 30

u = 18 cm

distance of object be equal to u = 18 cm

6 0
3 years ago
a solution made by dissolving 116 g of CaCl2 in 64 g of water has a density of 1.180 g/ml at 20 degrees celsius what is the mola
Makovka662 [10]

Answer:

The molarity of CaCl2 in the solution is 4.94 M

Explanation:

First of all you need to calculate CaCl2 mass.

You have one atom of Ca = 40.07 g/mol and two atoms of Cl = 35.45 g/mol so the molecule has a mass of 110.97 g/mol.

Now, knowing that your solution will have a mass of 64 grams of water + 116 grams of CaCl2 = 180 grams, you can calculate its volume, knowing that density = mass/volume

density x mass = volume --> 1,180 g/ml x 180 g = 212.4 ml

In 212.4 ml, you have 116 grams of CaCl2. You can calculate how many moles of CaCl2 you have:

110.97 g ------ 1 mol

116 g -------- x = (116 g x 1 mol) / 110.97 g =  1.05 moles

The molarity in a solution equals how many moles of a certain solute you have in 1000 ml of solution. In this solution, you have 1.05 moles in 212.4 ml, so in 1000 ml you will have:

212.4 ml ------- 1.05 moles

1000 ml -------- x = (1000 ml x 1.05 moles) / 212.4 ml = 4.94 moles.

This means the molarity of CaCl2 in the solution is 4.94 M.

4 0
4 years ago
If a voltmeter has a less than ideal resistance, say 1 MΩ, and is used to measure the voltage across a resistor of a comparable
Naddik [55]

Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

lets suppose the supplied voltage is 1 V.

First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

R = 0.33 MΩ

As you can see the voltmeter will see a reduced resistance and the corresponding voltage also reduces because resistance and voltage are directly proportional.

Therefore, it is preferred to have a very high parallel resistance of the voltmeter.

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cricket20 [7]

Answer:

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Explanation:

Hope this helps!!!!

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galben [10]

Answer:

the second one!

Explanation:

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4 0
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