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alisha [4.7K]
3 years ago
8

A local AM radio station broadcasts at a frequency of 696 KHz. Calculate the energy of the frequency at which it is broadcasting

.
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

E=4.61\times 10^{-28}\ J

Explanation:

Given that,

The frequency of local AM radio station, f = 696 KHz = 696000 Hz

We need to find the energy of the frequency at which it is broadcasting.

We know that,

Energy of a wave, E = hf

Where

h is Planck's constant

Put all the values,

E=6.63\times 10^{-34}\times 696000\\\\=4.61\times 10^{-28}\ J

So, the energy of the wave is equal to4.61\times 10^{-28}\ J.

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An electron in an atom's orbital shell, labeled X in the model below, released enough energy to move to a different orbital shel
Delicious77 [7]

Answer:

Lower energy shell which will be nearer to the nucleus.

Explanation:

When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.

When atoms releases energy, electrons move into lower energy levels.  The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.

Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away

7 0
3 years ago
Water drips from the nozzle of a shower onto the floor 190 cm below. The drops fall at regular (equal) intervals of time, the fi
VARVARA [1.3K]

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

t = \sqrt{0.388} = 0.62 s

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

8 0
3 years ago
A 9V battery is directly connected to each of 3 LED bulbs. Select the statement that accurately describes this circuit.
dem82 [27]

Answer:

where are the statements?

Explanation:

8 0
3 years ago
Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
4 years ago
A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the t
oee [108]

Answer:

v_x = 1.26 m/s

Explanation:

given,

weight of swimmer = 510 N

length of ledge, L = 1.75 m

vertical height of the cliff, h =  9 m

speed of the swimmer = ?

horizontal velocity  of the swimmer should be that much it can cross the wedge.

distance = speed x time

d = v_x × t

1.75 = v_x × t ........(1)

now,time taken by the swimmer to cover 9 m

initial vertical velocity of the swimmer is zero.

using equation of motion for time calculation

s = ut +\dfrac{1}{2}gt^2

9= 0+\dfrac{1}{2}\times 9.8\times t^2

  t² = 1.938

  t = 1.39 s

same time will be taken to cover horizontal distance.

now, from equation 1

1.75 = v_x × 1.39

v_x = 1.26 m/s

horizontal speed of the swimmer is equal to 1.26 m/s

4 0
3 years ago
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