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ASHA 777 [7]
4 years ago
12

What is a process that returns to its beginning and repeat it self in the same sequence

Physics
1 answer:
stepladder [879]4 years ago
8 0

It is also known as iteration. It is used mostly in computer programming with for and for each loops in which iteration is present. In mathematics there are some "interational operations" like product and sum for eg:

J=\sum_{i=1}^{n}J_i\Longrightarrow J_1+J_2+\dots+J_{\infty}

The above equation describes the summation of J alike variables n times.

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2 years ago
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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
What is a law in physics? state the criteria for a statement to be a law.​
steposvetlana [31]

Answer:

By nature, laws of Physics are stated facts which have been deduced and derived based on empirical observations. Simply put, the world around us works in a certain way, and physical laws are a way of classifying that “working.”

8 0
3 years ago
When light is directed on a metal surface, the kinetic energies of the photoelectrons a) are random b) vary with the frequency o
jekas [21]

Answer:

b) vary with the frequency of the light

Explanation:

The phone electric effect can be expressed as

K.E=(hv -W•)

Where K.E is the Kinectic energy

W• = work function of the metal

ν =frequency of the radiation

h = Planck's constat

Then, we can see that K.E is proportional linearly to "v" in the equation above.

Therefore, When light is directed on a metal surface, the kinetic energies of the photoelectrons vary with the frequency of the light

5 0
3 years ago
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

5 0
4 years ago
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