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4 years ago

Given the quantities a= 9.7 m, b= 4.2 s, c= 69 m/s, what is the value of the quantity d = a^3/(cb^2)?

Physics

2 answers:

nexus9112 [7]4 years ago

7 0

D= 9.7^3/(69)(4.2)^2
d=912.673/289.8^2
d=912.673/83984.04
d=0.01086721953362...
I hope this helped!

juin [17]4 years ago

3 0

Answer = a^3 / c.b^2

A = (9,7m)^(3) / 69m/s×(4,2s)^(2)

A = 912,673m^(3)/ 69m/s ×17,64s^2

A = 912,673m^(3) / 1.217,16m.s

A = 0,749838147 m^(2)/s

Or

A = 7,5 × 10^(-1) m^(2)/s

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Hi there!

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Answer:

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Y component of magnetic field is -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

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$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

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(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

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