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3 years ago

An electron in a mercury atom changes from energy level b to a higher energy level when the

Physics

1 answer:

emmainna [20.7K]3 years ago

4 0

The energy is 3.06 electronvolts, E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J

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In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0

2 years ago

A particular lightbulb is designed to consume 40 W when operating on a car's 12-V DC electric power. If you supply that bulb wit

Andrew [12]

40V because it will provide the same amount of power.

4 0

2 years ago

A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t

andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be ( R+ h )

orbital velocity V₀ = $\sqrt{\frac{GM}{( R+h)} }$

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x $\sqrt{\frac{( R+h)}{GM } }$

$\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ = R +h

h = $\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ - R.

7 0

3 years ago

Read 2 more answers

The process by which plastic deformation is produced by dislocation motion is called

QveST [7]

Answer:gock uwu

Explanation:

3 0

2 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago