PLEASE HELP ME WITH THIS ONE QUESTION

Physics

1 answer:

melamori03 [73]3 years ago

3 0

Explanation:

First, we convert the energy from eV to Joules:

$2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)$

$= 4.64×10^{-19}\:\text{eV}$

We know from definition that

$E=h\nu = \dfrac{hc}{\lambda}$

so the wavelength of the photon is

$\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}$

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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i

3 0

4 years ago

Read 2 more answers

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$