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2 years ago

A temperature of a 0.349 kg sample of copper decreases from 87.0 °C to 21.0 °C. How much heat

Physics

1 answer:

Arte-miy333 [17]2 years ago

7 0

We have that heat flow out of the copper sample is mathematically given as

Q=8.86809J

Question Parameters

A temperature of a 0.349 kg sample of copper
decreases from 87.0 °C to 21.0 °C.

Endothermic Reaction

It is common knowledge that an endothermic reaction is one where the system absorbs heat energy from the <em>environs</em>.

Exothermic Reaction

An exothermic reaction is one where the system releases heat energy to the <em>environs</em>.

Generally the equation for Heat is mathematically given as

$Q=mCdt\\\\Therefore\\\\Q=0.349*0.385*(87-21)$

Q=8.86809J

For more information on temperature visit

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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

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

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

Q1Q2 = 1.49 × $10^{-9}$ C²

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 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

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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3 years ago

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Answer:

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Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

$\Theta=\frac {\pi}{2}$

Speed of running of women =5 m/s.

Time of running = 5 minutes= 5 x 60 secondes= 300 s.

(a) As distance= (speed) x (time)

So, the required distance= 5 x 300 m= 1500 m.

(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

As arc length, $l= \theta R$ , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.

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In 1 cycle, there are 2 swings, so distance covered in 1 cycle $= 3 \pi m$ .

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Answer:

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