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Sloan [31]
3 years ago
7

Assertion: When I P+ Q I = I P- QI, then P must be perpendicular to Q. Reason: The relation will hold even when Q is a null vect

or.
Physics
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

The assertion is true and reason is false.

Explanation:

Assertion: I P+ Q I = I P- QI, then P must be perpendicular to Q.

Reason : The relation will hold even when Q is a null vector.

Now

\left | P + Q \right |=\left | P - Q \right |\\\\P^2 + Q^2 + 2 P Q cos \theta =P^2 + Q^2 - 2 P Q cos \theta\\\\4 P Q cos \theta = 0 \\\\cos \theta = 0 \\\\\theta = 90 degree

So, P and Q are perpendicular to each other.

So, the assertion is true.

Reason is false.

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The first law of thermodynamics can be given as ________.
MrRa [10]
The first law of thermodynamics can be written as
\Delta U = Q-W
where
\Delta U is the variation of internal energy of the system
Q is the amount of heat absorbed by the system
W is the work done by the system on the surrounding.

Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
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7 0
3 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

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Answer:

Explanation:

energy friction

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I uploaded the answer t^{}o a file hosting. Here's link:

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