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OLga [1]
3 years ago
14

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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Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>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 </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

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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

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
Which variable is changed in an experiment?
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C. The Independent variable

It is the variable that you manipulate, while dependent is the response.
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3 years ago
A weightlifter lifts a weight of 500 N from the ground over her head, a distance of 1.8 m. How much work has been done to move t
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Answer is 900N

Hope this helps c:
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3 years ago
The gravitational potential is a distance z=0.17 m away from a distribution of mass that has the following equation:
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gravitational potential is given as

V = \frac{GM}{R^2}(R^2 + Z^2)^0.5

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E  = -\frac{GM}{R^2}*0.5(R^2+z^2)^-0.5*2z

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M = 110 kg

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E = -\frac{6.67 * 10^{-11}* 110*0.17}{0.55^2*(0.55^2 + 0.17^2)^0.5}

E = 7.16 * 10^{-9} N/kg

so above is the field intensity

4 0
3 years ago
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