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2 years ago

For an isolated system, the sum of the kinetic and potential energies

Physics

2 answers:

pickupchik [31]2 years ago

5 0

I think it B I hope this help

Elenna [48]2 years ago

4 0

I am pretty sure the answer to your question is B

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What is the natural frequency (wo) of a circuit that has an inductor of 0.048 henery's, and a capacitor of 0.0004 farads?

adell [148]

Answer:

The natural frequency will be 228.11 rad/sec

Explanation:

We have given Inductance L = 0.048 Henry

And the capacitance C = 0.0004 farad

We have to find natural frequency

When only inductor and capacitor is present in the circuit then it is known as LC circuit and Natural frequency of LC circuit is given by $\omega _0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.048\times 0.0004}}=228.21\ rad/sec$

So the natural frequency will be 228.21

4 0

2 years ago

A baseball weighing 0.5 kg falls from the sky. You hit it with a bat with a force of 50 N at an upward angle of 45°.

kirill [66]

Answer:

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hjoufu I'm not sure if you are still interested in the position and would like to know if you are interested

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3 years ago

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$