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kolezko [41]
4 years ago
12

B. CO

Physics
2 answers:
jekas [21]4 years ago
6 0
The answer is 0.6 ms
Hunter-Best [27]4 years ago
5 0

Answer:

0.6 m/s

Explanation:

2Hz = 2^-1 = 2 /s

30cm = .3m

Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.

(.3m)*(2 /s) = 0.6 m/s

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15. What is one way in which scientists study dark
Kay [80]

Answer:

It's d

Explanation:

Scientists study dark matter by looking at how it affects visible matter.

8 0
3 years ago
_____ can be reduced but it cannot be eliminated.
Sonbull [250]
Heat can be reduced but it cannot be eliminated completely there is some amount if heat energy present even if it is absolute zero(-273degrees Celsius)
3 0
4 years ago
Read 2 more answers
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

6 0
4 years ago
In the Electromagnetic Force, opposite charges will ________.
kiruha [24]
-- Charges with the same sign repel each other.

-- Charges with different signs attract each other.

-- This is called the "electrostatic force".

-- There's another set of forces called "magnetic forces", but

-- there's no such thing as "electromagnetic force".
6 0
3 years ago
Read 2 more answers
Very short pulses of high-intensity laser beams are used to repairdetached portions of the retina of the eye. The brief pulses o
maw [93]

Answer:

a)  W = 3.87 10⁻⁴ J , b)  P = 3.10 10⁻³ Pa , c)  λ = 671. 6 nm , d) frequency does not change , e)  Emax = 1.39 C / m  and f) Bmax = 4.7 10⁻⁹ T

Explanation:

a) Let's use the concepts of power that is work for the unit of time and work is the change of kinetic energy

    P = W / t

An electromagnetic wave has an intensity

    I = P / A =

   W = P t

   W = 258 10⁻³ 1.5 10⁻³

   W = 3.87 10⁻⁴ W s

   W = 3.87 10⁻⁴ J

b) the radiation pressure is given by the ratio

    P = I / c

Where I is the intensity

     I = Powers / A

    A = π r² = π (d/2)²

     I = 258 10⁻³ / π (297.5 10⁻⁶)²

     I = 9.29 10 5 W / m²

     P = 9.29 10⁵/3 10⁸

     P = 3.10 10⁻³ Pa

c) The wavelength when passing a measured of different refractive index changes in the way

     λ = λ₀ / n

     λ = 900 10⁻⁹ / 1.34

     λ = 671.6 10⁻⁹ m

     λ = 671. 6 nm

d) when the light strikes a medium creates a forced oscillation in the electrons of the medium, this is a resonance phenomenon, so the frequency does not change

e) The maximum electric field is

      I = Emax2 / 2 μ₀ c

     Emax = Ra (2 μ₀ c I)

     Emax = Ra (2 4 π 10⁻⁷ 3 10⁸ 258 10⁻³)

     Emax = 1.39 C / m

f) the elective and magnetic fields are related

     c = Emax Bmax

     Bmax = Emax / c

     Bmax = 1.39 / 3 10⁸

     Bmax = 4.7 10⁻⁹ T

7 0
3 years ago
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