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sp2606 [1]
3 years ago
15

Un conductor se conecta en posición horizontal cargado positivamente con una densidad lineal de carga de 12mc/m ( 12 e-6 C)y a u

na cierta distancia del conductor se ubica una esfera de sauco de una masa de 3mg, presentando una carga de 18mc ¿ a qué distancia debe estar la esferita para que se mantenga el equilibrio Un conductor se conecta en posición horizontal cargado positivamente con una densidad lineal de carga de 12mc/m y a una cierta distancia del conductor se ubica una esfera de sauco de una masa de 3mg, presentando una carga de una masa de 18mc ¿ a qué distancia debe estar la esferita para que se mantenga el equilibrio estático?
Physics
1 answer:
viktelen [127]3 years ago
7 0

Answer:

ok so u 4394 and 2340234 dhw04436445 ?! there

Explanation:

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What is the velocity of a car that traveled a total of 75 kilometers north in 1.5 hours?
swat32

The car's velocity is (distance + direction) / (time) =

                                 (75 km-north) / (1.5 hrs)  =

                                 (75/1.5)  (km-north/hr)  =  50 km/hr  north.

7 0
3 years ago
The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that
OLEGan [10]

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

=80dB

6 0
3 years ago
If a system has 475 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in int
Andrews [41]
First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ

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ΔU = -500 + 1987.4
ΔU = 1487.4 kJ
4 0
3 years ago
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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
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<span>The angular momentum of a particle in orbit is 

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Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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3 0
3 years ago
LOTS OF BRAINLIST WILL BE GIVING TO THOSE WHO HELP
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You know you can skip those and just submit them, they don’t even check them
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