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SSSSS [86.1K]
3 years ago
5

This is a substance that cannot be broken down further by chemical means.

Physics
2 answers:
viktelen [127]3 years ago
5 0

The answer is ELEMENT

Serggg [28]3 years ago
4 0
I think its an element
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Explanation:

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A uniformly charged thin ring has radius 15.0 cm and total charge 20.0 nC . An electron is placed on the ring's axis a distance
Ivanshal [37]

Answer:

b) 1.67×10^7 m/s

Explanation:

The solution is attached in the attachment section

8 0
3 years ago
Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a
Yuliya22 [10]

Answer:

The value is q = 3.4 *10^{-6} \ C

Explanation:

From the question we are told that

    The mass of each sphere is m_1 = m_2  = m  =  0.020 \ kg

     The length of the string is  l = 0.18 \  m

     The angle of with the vertical is \theta  =  7^o

      The acceleration due to gravity is g = 9.8 \ m/s^2

Generally the force acting between the forces is mathematically represented as

       F  =  T cos \theta =  \frac{k*  q^2}{ r^2}

=>     T cos \theta =  \frac{k*  q^2}{ r^2}

Generally from Pythagoras theorem the radius of the circular curve created by the force is

         r = 2 L sin (\theta )

=>      r = 2* 0.180 sin (7)

=>      r = 0.043 \ m  

=>     q = tan \theta * \frac{m * g * r^2 }{k}

=>      q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}

=>      q = 3.4 *10^{-6} \ C

7 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
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