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3 years ago

30pts!!!06.01 Simple Harmonic Motion Lab: Virtual Lab Simulation

Physics

2 answers:

Ostrovityanka [42]3 years ago

8 0

Answer:

Hello; I would like to say you that brainly is for short answers not very long answers if you want the answer you can contact brainly services

mel-nik [20]3 years ago

8 0

Answer:

https://prezi.com/020q6srszjtp/simple-harmonic-motion-lab/

Explanation:

i found a prezi slide show for this online with answers and explanations :)

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A gas is confined in a cylinder with a piston. What happens when work is done on the gas.

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It combusts in to a vapor

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A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara

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Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

$r=2.3*sin70=2.161m$ (ball rotates in a circle)

If the system is in equilibrium, the tension is:

$Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}$

Replacing:

$\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}$

Replacing:

$v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s$

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3 years ago

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3 years ago

Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S an

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Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

$F=k_e\frac{q_1q_2}{r^2}$

where,

F = electrostatic force of attraction or repulsion

$k_e$ = Coulomb's constant

$q_1$ and $q_2$ are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

$F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2$ ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

$F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}$ ...........(2)

Equation equation 1 and 2, we get

$\frac{F_{sq}}{F_{sp}}=\frac{1}{4}$

$F_{sq}=4\times F_{sp}$

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

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