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3 years ago

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Matt forgot to put the fabric softener in the wash. As his socks tumbled in the dryer, they became charged. If a small piece of

lint with a charge of +1.25 E -19 C is attracted to the socks by a force of 3.0 E -9 N, what is the magnitude of the electric field at this location?

1 answer:

mr_godi [17]3 years ago

5 0

Answer:

$2.4\cdot 10^{10} N/C$

Explanation:

The electric force exerted on a charge due to an electric field is given by:

$F=qE$

where

q is the magnitude of the charge

F is the magnitude of the force

E is the magnitude of the electric field

In this problem, we have:

$q=1.25\cdot 10^{-19}C$ is the charge

$F=3.0\cdot 10^{-9} N$ is the magnitude of the force

using these data and re-arranging the equation, we can calculate the magnitude of the electric field:

$E=\frac{F}{q}=\frac{3.0\cdot 10^{-9} N}{1.25\cdot 10^{-19} C}=2.4\cdot 10^{10} N/C$

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Answer:

<h3>The answer is option A</h3>

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3 years ago

Read 2 more answers

Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

fomenos

Answer:

(a) $\vec{R}= 4.83\ m\ \hat{i}+1.47\ m\ \hat{j}$

(b) (5.05 m, 16.93 degrees wrt x-axis)

Explanation:

Given:

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$\vec{E}$ = (4.50 m, 53.0 degrees wrt x-axis)

Let us first fond out vector D and E in their rectangular form.

$\vec{D} = (3\cos 315^\circ\ \hat{i}+3\sin 315^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.12\ \hat{i}-2.12\ \hat{j})\ m\\$

Similarly,

$\vec{E} = (4.5\cos 53^\circ\ \hat{i}+4.5\sin 53^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.71\ \hat{i}+3.59\ \hat{j})\ m\\\because \vec{R}=\vec{D}+\vec{E}\\\therefore \vec{R} = (2.12\ \hat{i}-2.12\ \hat{j})\ m+(2.71\ \hat{i}+3.59\ \hat{j})\ m\\\Rightarrow \vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

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We can write the resultant vector R as below:

$\vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (b):

$Magnitude\ of\ resultant = \sqrt{4.83^2+1.47^2}\ m = 5.05\ m\\\textrm{Direction in angle with the x-axis} = \theta = \tan^{-1}(\dfrac{1.47}{4.83})= 16.93^\circ$

Since both the components of the resultant lie on the positive x and y axes. So, the resultant makes an acute angle with the positive x-axis.

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3 years ago

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