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disa [49]
3 years ago
14

Gggcdfubrdegubtcwftvf y day rx u e HHS’s forget h

Physics
2 answers:
aev [14]3 years ago
8 0

Answer:

um . . . yes ?

Anika [276]3 years ago
4 0

True, but tditdu4siseuezirsrsuesywsuesuesues7 eyes 7esuesuesuexiyfyplyfotdirsirsltsottif

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Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3
Mekhanik [1.2K]

Answer:

Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N

F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}

The net force on q3 is the sum of these two forces:

F_{net} = F_{13} + F_{23} = -1.92\times 10^{-5} + (-19.5\times10^{-6}) = 2.8\times 10^{-5} N

Since both forces are directed towards left, their sign should be negative.

5 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
3 years ago
Three noise sources produce volume (loudness) levels of 70, 73, and 80 dB when acting separately. When the sources act together,
Mashutka [201]

Sound at 70 dB is 70 dB louder than the human reference level.  That's 10⁷ times as much as the reference sound power.

Sound at 73 dB is 73 dB louder than the human reference level.  That's 10⁷.³  or  2 x 10⁷  times as much as the reference sound power.

Sound at 80 dB is 80 dB louder than the human reference level.  That's 10⁸  or 10 x 10⁷ times as much as the reference sound power.

Now we can adumup:

Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)

Intensity = (13 x 10⁷) times the sound power reference intensity.

Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)

Intensity = 70 + 10 log(13)

Intensity = 70 + 10 (1.114)

Intensity = 70 + 11.14

Intensity = <em>81.14 dB</em>

<em>______________________________________</em>

Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct.  For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.

8 0
3 years ago
You have a nightlight plugged into an outlet in the hallway, which uses 3.5 watts when plugged in. If the house circuit provides
nekit [7.7K]
Base on the said question or problem that state and ask to calculate the current of the said light bulb and in my further calculation and further analysis, I would say that the current of the light bulb would be 0.0292. I hope you are satisfied with my answer and feel free to ask for more 
7 0
4 years ago
Read 2 more answers
In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood
aksik [14]

Answer:

  • Here we use the conservation of momentum theorem.

  • m stands for mass, and v stands for velocity. The numbers refer to the respective objects.

  • m1v1 + m2v2 = m1vf1 + m2vf2

  • Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.

  • m1v1 + m2v2 = vf(m1 + m2)

<u>Let’s substitute in our givens.</u>

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

3 0
3 years ago
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