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a_sh-v [17]
4 years ago
12

How many electrons must be transferred from an object to produce a charge of 1.15 c? (express your answer to three significant f

igures.)?
Physics
1 answer:
Darina [25.2K]4 years ago
6 0
<span>To produce a charge of 1.15 Coulombs in an object, we need to transfer = 1.15/(charge on an electron) number of electrons

   As we know that, total charge on an electron = 1.60217662 Ă— 10 ^-19 coulombs
    Hence, number of electrons required to move = 1.15/(1.60217662 Ă— 10 ^-19) = 0.71777 x 10^19 electrons, i.e. 7.17 x 10^18 electrons upto 3 significant figures</span>
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EDIT: He has these backwards, the shortest wavelength is created by Gamma-Rays and the longest is Radiowaves.

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A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
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A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
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k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
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since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
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v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

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U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
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