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Vladimir [108]
3 years ago
15

suppose two masses are connected by a spring. compute the formula for the trajectory of the center of mass of the two mass oscil

lator
Physics
1 answer:
Ymorist [56]3 years ago
8 0

Answer:

The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)

Explanation:

let the first mass = M

let the position of second  = M

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. Inside a conducting sphere of radius 1.2 m, there is a spherical cavity of radius 0.8 m. At the center of the cavity is a poin
MAVERICK [17]

Answer:

 E = 1,873 10³ N / C

Explanation:

For this exercise we can use Gauss's law

        Ф = E. dA = q_{int} / ε₀

Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.

The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.

        The surface of a sphere is

             A = 4π r²

             E 4π r² = q_{int} /ε₀

  The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is

           q_{int} = q₁ + q₂

           q_{int} = (530 - 200) 10⁻⁹

           q_{int} = 330 10⁻⁹ C

The electric field is

             E = 1 / 4πε₀   q_{int} / r²

            k = 1 / 4πε₀

            E = k q_{int}/ r²

Let's calculate

           E = 8.99 10⁹   330 10⁻⁹/ 1.32²

           E = 1,873 10³ N / C

7 0
3 years ago
A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the object's density in units of g/ml?
Zinaida [17]
G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
2.20 cm, 1.35 cm, and 1.25 cm
Then the total volume is: V = lwh = 3.7125 cm^3
To get the density, we divide mass by volume: 2.50 g / 3.7125 cm^3 = 0.6734 g/cm^3 = 0.6734 g/mL
4 0
3 years ago
The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscop
Blababa [14]

Answer:

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

   =18/2.76

fₙ =  6.52 cm

The focal length of eye piece is 6.52 cm.

6 0
3 years ago
Please HELP ASAP PLEASE I WILL MARK BRAINLIEST
Anika [276]
It’s True the volume is 1k of iron is = 1
3 0
3 years ago
Pls help i begg youuuuu
Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
<span>
3. 343m/s / 0.65552m = 523.349 Hz
    </span>This corresponds to pitch C

4. using f = c /  λ
  λ = c / f<span>
     = 343m/s / 587.33 = 0.583999 m = 0.584 m

</span>
3 0
3 years ago
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