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leva [86]
3 years ago
5

If a drop is to be deflected a distance d = 0.350 mm by the time it reaches the end of the deflection plate, what magnitude of c

harge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m3 , and ignore the effects of gravity.
Physics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

q = 4.87 X 10^ -14 C

Explanation:

As d=0.350 mm

The ink drop will be accelerated by the electric field between the plates:

a = F/m

d = a(D0 / v)^2 / 2 ...... 1

a = qE/m ............... 2

Substituting 2  into 1:

d = (qE/m)(D0 / v)^2 / 2

q = 2mdv^2 / [E(D0)^2]

q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]

q = 4.87e-14 C

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Answer:

θ = 30⁰  

Explanation:

We have been given two wave displacement equations,

y_{1} =2sin\theta +1     ........(1)

y_{2}=3sin\theta +0.5     ........(2)

The waves will have same displacements when y_{1}=y_{2}=y (say)

Therefore, operating (2) - (1), we get,

sin\theta - 0.5=0

or, \theta = sin^{-1}(0.5)=30^{o} (since, sin 30⁰ = 0.5)

We can check the answer by putting \theta=30^{o} in equations (1) and (2),

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3 years ago
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Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
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Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

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Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

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E=(30\times10^{-3}+5.29)\times 9.8\times0.15

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A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction
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Given:

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let  '\theta' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

cos\theta = \frac{1\times 1 + 1\times 0 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+0^{2}+1^{2}}}}

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let  '\phi' be the angle between [1 1 1] and [1  1 1], then by coordinate formula:

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\sigma _{1} = \frac{T_{cRss}}{cos\theta  cos\phi }

\sigma _{1} = \frac{0.242\times 10^{6}}{\sqrt{\frac{2}{3}}\times 1}

\sigma _{1} = 0.2964 MPa

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3 years ago
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