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navik [9.2K]
3 years ago
10

A 0.13 g plastic bead is charged by the addition of 1.0Ã1010 excess electrons. what electric field eâ (strength) will cause the

bead to hang suspended in the air?
Physics
1 answer:
slavikrds [6]3 years ago
5 0
The working equation for this is: E = F/Q, where E is the strength of electric field, F is the force and Q is the charge. The force is equal to:

F = mg = (0.13/1000 kg)(9.81 m/s²) = 1.2753×10⁻³ N

The charge of he excess electrons is equal to:
Q = (-1.6021766208×10⁻¹⁹ C/electron)(1×10¹⁰ electrons)
Q = -1.6021766208×10⁻⁹ C

E = 1.2753×10⁻³ N/-1.6021766208×10⁻⁹ C
E = -795,979.66 N/C
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Answer:

1185 N

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Net force acting on motorcycle F_{net} is given by  F_{net}=F-f

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for F_{net}

F_{net}=ma hence  ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

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In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
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Explanation:

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d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

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(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

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\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

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8 0
3 years ago
Find the measure of angle x in the figure below:
Serga [27]

Answer:

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4 0
3 years ago
A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm
Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

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The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

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we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0
3 years ago
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