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ExtremeBDS [4]
3 years ago
15

If a photon has a frequency of 5.20 × 1014 hertz, what is the energy of the photon? Given: Planck's constant is 6.63 × 10-34 jou

le·seconds.
Physics
1 answer:
Virty [35]3 years ago
4 0
Answer:
E=h.f

=5.2×1014×6.63×10^-34

E=3.496×10^-30 J
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Assume that charge −q−q-q is placed on the top plate, and +q+q+q is placed on the bottom plate. What is the magnitude of the ele
STatiana [176]

Answer:

Magnitude of electric field = E = q/Aε0

Explanation:

Consider plates are placed at a distance of d. As given in the question the charge stored on the plates have magnitude q and given by:

                                          q = CV

And  

                                          V = q/C    ……. (i)

The capacitance is given by the following equation:  

                                         C = Aε0/d ……. (ii)

Put equation (ii) in (i) ,

                                          V = qd/ Aε0 …..(iii)    

The electric field is defined as:  

                                            E = V/d   …… (iv)

Put equation (iii) in (iv),

                                            E = qd/ Aε0d

                                            E = q/Aε0

Hence, the magnitude of electric field will be q/Aε0 .

                                         

8 0
3 years ago
Can someone help me with this Physics question please?
strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

N=N_0(\frac{1}{2})^{\frac{t}{H} where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

N=48.0(\frac{1}{2})^{\frac{49.2}{12.3} and simplifying a bit:

N=48.0(.5)^4 and

N = 48.0(.0625) so

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6 0
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Explanation:

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Annette [7]

Answer: i think its local drive

Explanation:

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