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koban [17]
3 years ago
10

A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective dura

tion of the flash is 0.320 s, during which it produces an average 0.620 W from an average 3.00 V.
a. How much charge moves through the lamp (C)?b. find the capacitance (F)c. What is the resitance of the lamo?
Physics
1 answer:
Genrish500 [490]3 years ago
6 0

Power of the lamp = (voltage) x (current)

0.62 watt = (3.0 v) x (current)

Current = (0.62 watt) / (3.0 volt)

Current = 0.207 Ampere

Charge = (current) x (time)

Charge = (0.207 Amp) x (0.32 sec)

<em>a).  Charge = 0.066 Coulomb</em>

Energy discharge through the lamp = (charge) x (voltage)

Energy = (0.66 C) x (3.0 volt)

Energy = 0.1984 Joule

Energy stored in a capacitor = (1/2) (Capacitance) (voltage²)

0.1984 Joule = (1/2) (Cap) (3.0 v)²

Cap = 2 x 0.1984 / 9

<em>b).  Cap = 0.044 Farad</em> (yikes !)

Resistance of the lamo = (voltage) / (current)

Resistance = (3.0 volt) / (0.207 Amp)

<em>c).  Resistance = 14.49 Ω</em>

All of this feels OK except for that ridiculous 44,000 μF, 3v capacitor.  Sadly, I have checked my work without finding an error, and I am too lazy to check it again.  So I'll call this my final answer, and leave it to some better Brainly person to find where I went wrong.

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The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string
Katyanochek1 [597]

Answer:

\dfrac{\mu_A}{\mu_G}=0.197

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G  = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

 v_A = \sqrt{\dfrac{F_A}{\mu_A}}.......(1)

speed for string G

 v_G = \sqrt{\dfrac{F_G}{\mu_G}}........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{\mu_A}{\mu_G}=0.197

5 0
2 years ago
Which statement about a right triangle is NOT true?
Alex787 [66]

Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two lengths.

Explanation:

only one that made sense

6 0
3 years ago
Read 2 more answers
Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani
Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

n = \frac{t}{t_{1/2}}

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

t_{1/2} = half-life = 700 million years

Therefore,

n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3

Now, we will calculate the number of uranium nuclei left (n_u):

n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u =  800\ million

and the rest of the uranium nuclei will become thorium nuclei (u_{th})

n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million

dividing both:

\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

6 0
2 years ago
Question 7 of 10
Dahasolnce [82]
Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion.
https://www.khanacademy.org › stat...
Static and kinetic friction example (video) | Khan Academy

Answer a would be static friction
Answer b is fluid friction
(Air resistance is fluid friction. Fluid friction is the friction experienced by objects which are moving in a fluid and the air is a fluid.)
Answer c is static friction
ANSWER D IS KINETIC FRICTION

Hope this helps :D



4 0
2 years ago
A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.
Murrr4er [49]

Answer:

Hi how are you doing today Jasmine

8 0
2 years ago
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