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3 years ago

ANSWER ASAP

Physics

2 answers:

lidiya [134]3 years ago

5 0

Answer: The answer is B.

Please mark me brainiest

vodka [1.7K]3 years ago

3 0

B the desert was once covered in water

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Just as in a bar magnet, the magnetic field of a wire with a current forms a cylinder shape around the wire.

Ne4ueva [31]

Yes, the magnetic field of a wire carrying a current forms a cylinder shape around the wire, but that's not similar to the field of a bar magnet.

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3 years ago

What happens when an electron moving from the 3rd energy level to the 1st energy level?

Sholpan [36]

Answer:

A photon of wavelength 103 nm is released

Explanation:

When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.

For example, if we are talking about a hydrogen atom, the energy of the levels are:

$E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV$

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

$\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV$

In Joules,

$\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J$

We can also find the wavelength of this photon, using the equation:

$\Delta E = \frac{hc}{\lambda}\rightarrow \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm$

7 0

4 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

CH, +20, → CO2 + 2H,0 CH, +20, are the of the above chemical reaction.

rjkz [21]

These are the reactants because they are on the left side of the equation.

7 0

3 years ago

In a nuclear power plant, _____. energy is released from the nuclei of atoms energy is released from the bonds of molecules ener

oksian1 [2.3K]

In a nuclear power plant, energy is released from the nuclei of atoms. The correct option among all the options given in the question is the first option. Huge amount of thermal energy is released by the breaking of the uranium atoms. This energy is used for turning a turbine that produces electricity. It is a very clean method of producing electricity.

8 0

3 years ago

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