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1 year ago

Calculate the volume of material whose density is 0.798gcm-3 and a mass 25g

Physics

2 answers:

luda_lava [24]1 year ago

5 0

Explanation:

density equals mass/volume

Elena L [17]1 year ago

4 0

<u>Answer:</u>

$\mathrm{volume = \bf 31.33 \sapce\ cm^{-3}}$

<u>Explanation:</u>

To calculate the volume, we can use the following formula:

$\boxed{\mathrm{density = \frac{mass}{volume}}}$ .

In this case,

• density = 0.798 gcm⁻³

• mass = 25 g

Substituting values into the formula:

$\mathrm{0.798 \space\ gcm^{-3}= \frac{25 \space\ g}{volume}}$

⇒ $\mathrm{volume = \frac{25 \space\ g}{0.798 \space\ gcm^{-3}}}$

⇒ $\mathrm{volume = \bf 31.33 \sapce\ cm^{-3}}$

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Two small nonconducting spheres have a total charge of 93.0 μC . Part A

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Answer:

charge on each

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Q2 = 7.23 × $10^{-5}$ C

when force were attractive

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Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

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and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

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and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

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