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2 years ago

7

The normal formula to find force is F = m*a. What kind of math do you need to do

1 answer:

inessss [21]2 years ago

7 0

$\\ \sf\Rrightarrow F=ma$

Take a to left
As it's multiplied on right side it will be divided on right side.

$\\ \sf\Rrightarrow \dfrac{F}{a}=m$

Or

$\\ \sf\Rrightarrow m=\dfrac{F}{a}$

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Elements on the periodic table are grouped by their

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4 years ago

Read 2 more answers

Help please, I don't get it

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Answers:

a) $\hat F=(0.83,-0.55) N$

b) $\hat D=(-0.44,-0.89) m$

c) $\hat V=(-0.47,0.88) m/s$

Explanation:

A unit vector is a vector whose magnitude (length) is equal to 1. This kind of vector is identified as $\hat v$ and the way to calculate is as follows:

$\hat v=\frac{\vec v}{|v|}$

Where:

$\vec v=(x,y)$ is the vector

$|v|=\sqrt{x^{2}+y^{2}}$ is the magnitude of the vector

Having this information clarified, let's begin with the answers:

a) Force Vector

$\vec F=(9.0 \hat i - 6.0 \hat j) N$

Magnitude of $\vec F$ :

$|F|=\sqrt{(9.0 \hat i)^{2}+(-6.0 \hat j)^{2 }}N=10.81 N$

<u />

<u>Unit vector:</u>

$\hat F=\frac{\vec F}{|F|}$

$\hat F=\frac{(9.0 \hat i - 6.0 \hat j) N}{10.81 N}$

$\hat F=\frac{9.0}{10.81} N-\frac{6.0}{10.81}N$

$\hat F=(0.83,-0.55) N$

b) Displacement Vector

$\vec D=(-4.0 \hat i - 8.0 \hat j) m$

Magnitude of $\vec D$ :

$|D|=\sqrt{(-4.0 \hat i)^{2}+(-8.0 \hat j)^{2 }}m=8.94 m$

<u />

<u>Unit vector:</u>

$\hat D=\frac{\vec D}{|D|}$

$\hat D=\frac{(-4.0 \hat i - 8.0 \hat j) m}{8.94 m}$

$\hat D=\frac{-4.0}{8.94} Nm+\frac{-8.0}{8.94}m$

$\hat D=(-0.44,-0.89) m$

c) Velocity Vector

$\vec V=(-3.50 \hat i + 6.50 \hat j) m/s$

Magnitude of $\vec V$ :

$|V|=\sqrt{(-3.50 \hat i)^{2}+(6.50 \hat j)^{2}}m/s=7.38 m/s$

<u />

<u>Unit vector:</u>

$\hat V=\frac{\vec V}{|V|}$

$\hat V=\frac{(-3.50 \hat i +6.50 \hat j) m/s}{7.38 m/s}$

$\hat V=\frac{-3.50}{7.38} m/s+\frac{6.50}{7.38}m/s$

$\hat V=(-0.47,0.88) m/s$

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4 years ago

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) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms

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Explanation:

<em>Hey</em><em>,</em><em> </em><em>there</em><em>!</em>

<em>Here</em><em>,</em><em> </em><em>In</em><em> </em><em>question</em><em> </em><em>given</em><em> </em><em>that</em><em>, </em>

<em>potential</em><em> </em><em>difference</em><em> </em><em>(</em><em>V</em><em>)</em><em>=</em><em> </em><em>8</em><em>V</em>

<em>resistance</em><em> </em><em>(</em><em>R</em><em>)</em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>ohm</em>

<em>Now</em><em>,</em>

<em>According</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Ohm's</em><em> </em><em>law</em><em>,</em>

<em>V</em><em>=</em><em> </em><em>R</em><em>×</em><em>I</em><em> </em><em> </em><em> </em><em> </em><em>{</em><em> </em><em>where</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>current</em><em>}</em>

<em>or</em><em>,</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>V</em><em>/</em><em>R</em>

<em>or</em><em>,</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>8</em><em>/</em><em>1</em><em>0</em>

<em>Therefore</em><em>, </em><em> </em><em>current</em><em> </em><em>is</em><em> </em><em>4</em><em>/</em><em>5</em><em> </em><em>A</em><em> </em><em>or</em><em> </em><em>0</em><em>.</em><em>8</em><em> </em><em>A</em><em>.</em>

<em>(</em><em>A</em><em>=</em><em> </em><em>ampere</em><em> </em><em>=</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>current</em><em>)</em><em>.</em>

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em> </em>

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4 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

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