Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:

W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:

To convert this into joules, we use the conversion factor:

So, 
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
Answer:
Workdone = 2906.25J
Explanation:
Given the following data;
Force, F = 155
Distance, d = 0.75
Stroke, s = 25
Mathematically, work done is given by the formula;
Work done = force * distance * stroke
Where,
W is the work done
F represents the force acting on a body.
d represents the distance covered by the body.
s is the number of strokes.
Substituting into the equation, we have;
Workdone = 2906.25J
Answer:
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I assume you meant to say

Given that <em>x</em> = √3 and <em>x</em> = -√3 are roots of <em>f(x)</em>, this means that both <em>x</em> - √3 and <em>x</em> + √3, and hence their product <em>x</em> ² - 3, divides <em>f(x)</em> exactly and leaves no remainder.
Carry out the division:

To compute the quotient:
* 2<em>x</em> ⁴ = 2<em>x</em> ² • <em>x</em> ², and 2<em>x</em> ² (<em>x</em> ² - 3) = 2<em>x</em> ⁴ - 6<em>x</em> ²
Subtract this from the numerator to get a first remainder of
(2<em>x</em> ⁴ + 3<em>x</em> ³ - 5<em>x</em> ² - 9<em>x</em> - 3) - (2<em>x</em> ⁴ - 6<em>x</em> ²) = 3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3
* 3<em>x</em> ³ = 3<em>x</em> • <em>x</em> ², and 3<em>x</em> (<em>x</em> ² - 3) = 3<em>x</em> ³ - 9<em>x</em>
Subtract this from the remainder to get a new remainder of
(3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3) - (3<em>x</em> ³ - 9<em>x</em>) = <em>x</em> ² - 3
This last remainder is exactly divisible by <em>x</em> ² - 3, so we're left with 1. Putting everything together gives us the quotient,
2<em>x </em>² + 3<em>x</em> + 1
Factoring this result is easy:
2<em>x</em> ² + 3<em>x</em> + 1 = (2<em>x</em> + 1) (<em>x</em> + 1)
which has roots at <em>x</em> = -1/2 and <em>x</em> = -1, and these re the remaining zeroes of <em>f(x)</em>.
Answer:
1) Q_c = - ½ Q
, 2) Q_c = + ¼ Q
, 3) Q_c= 3/8 Q
Explanation:
For this exercise we must use that equal charges repel and charges with different signs attract. When two objects are in contact, the charges are evenly distributed between them.
Scenario 1.
when the two spheres touch the charge -Q is distributed between them, when separating each sphere has a charge = -1/2 Q
as there are no more interacts sphere C its charge is
Q_c = - ½ Q
Scenario 2
when the two spheres touch the charge -Q is distributed between them, when separating each one has a charge
Q_a = Q_d = - ½ Q
now the sphere D and C touch the charge is Q_net = -1/2 Q + Q = + ½ Q
when separating each sphere has half the charge
Q_d = Q_c = + ¼ Q
since sphere C has no more interaction, its charge is
Q_c = + ¼ Q
Scenario 3
A and B touch the net charge is Q_net = - Q + 0 = - Q
when parting
Q_a = Q_b = - ½ Q
now B and D touch, the charge is Q_net = - ½ Q +0 = - ½ Q
when parting
Q_b = Q_d = - ¼ Q
finally C and D touch
the net charge is Q_net = Q- ¼ Q = ¾ Q
when separating each one is left with half the load
Q_c = Q_d = 3/8 Q
Q_c= 3/8 Q