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vfiekz [6]
3 years ago
8

Four objects are situated along the y axis as follows: a 2.02 kg object is at 3.03 m, a 2.92 kg object is at 2.43 m, a 2.53 kg o

bject is at the origin, and a 4.02 kg object is at -0.501 m. What is the y-component of the center of mass of these objects
Physics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

0.975 m

Explanation:

The center of mass y = ∑m₁y₁/∑m₁ where m₁ = the individual masses and y₁ = the y - coordinates of the individual masses

So,

y = (2.02 kg × 3.03  m + 2.92 kg × 2.43 m + 2.53 kg × 0 m + 4.02 kg × -0.501 m)/(2.02 kg + 2.92 kg + 2.53 kg + 4.02 kg)

y = (6.1206 kgm + 7.0956 kgm + 0 kgm - 2.01402 kgm)/11.49 kg

y = 11.20218 kgm/11.49 kg

y = 0.975 m

So, the y- component of the center of mass of these objects is 0.975 m

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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

To learn more about Columb's law refer to the link;

brainly.com/question/1616890

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7 0
2 years ago
What is the angle of reflection of the incident angle is 35°?<br> 35<br> 25°<br> 55<br> 90
Anestetic [448]

Answer:

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4 0
3 years ago
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the
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Answer:

a)   m_v = m_s ((\frac{w_o}{w})² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

Explanation:

a) The angular velocity of a simple harmonic motion is

           w² = k / m

where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

silicon only

         w₉² = \frac{K}{m_s}

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         w² = \frac{k}{m_s + m_v}

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in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

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           m_v = m_s ((\frac{w_o}{w})² - 1)

b) let's calculate

          m_v = 2.13 10⁻¹⁶ [(\frac{20.4}{2.85})² - 1)]

          m_v = 1.07 10⁻¹⁴ g

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