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Ira Lisetskai [31]
3 years ago
6

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

ectric field at twice the distance from the point charge?
A) At twice the distance, the strength of the field is E/2.B) At twice the distance, the strength of the field is 2E.C) At twice the distance, the strength of the field is 4E.D) At twice the distance, the strength of the field remains equal to E.E) At twice the distance, the strength of the field is E/4.
Physics
1 answer:
Nana76 [90]3 years ago
7 0

Answer:

E

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice  the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

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Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5​
juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

or

T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}

Where

l is length of the pendulum

l\propto T^2

or

\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2 ....(1)

ATQ,

\dfrac{T_1}{T_2}=1.5

Put in equation (1)

\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0
2 years ago
A circular conical reservoir has depth 20 feet and radius of the top 10 feet. water is leaking out so that the surface is fallin
levacccp [35]

As per given condition we know that vertex angle of the cone is given as

tan\theta = \frac{R}{H}

so here we can say that vertex angle will remain constant

so here

\frac{r}{y} = \frac{R}{H}

\frac{r}{8} = \frac{10}{20}

r = 4 feet

now for the volume we can say

V = \frac{1}{3}\pi r^2 y

also we can say

r = \frac{y}{2}

so here we will have

V = \frac{1}{3}\pi (\frac{y}{2})^2y = \frac{1}{12}\pi y^3

now for volume flow rate

Q = \frac{dV}{dt} = \frac{3}{12}\pi y^2\frac{dy}{dt}

Q = \frac{1}{4}\pi y^2 v_y

now plug in all data

Q = \frac{1}{4} \pi (8)^2 (12) ft^3/h

Q = 603.2 ft^3/h

4 0
3 years ago
A ray of light exits from a material with a refractive index of 1.75, traveling into air. The angle of refraction is 25°. What w
Viktor [21]

Answer:

θi = 47.7°

Explanation:

The formula for the refractive index is as follows:

n = \frac{Sin\theta_i}{Sin\theta_r}

where,

n = refractive index = 1.75

θi = angle of incidence = ?

θr = angle of refraction = 25°

Therefore,

1.75 = \frac{Sin\ \theta_i}{Sin\ 25^o} \\\\(1.75)(Sin\ \ 25^o) = Sin\ \theta_i\\\\\theta_i = Sin^{-1}(0.739)

<u>θi = 47.7°</u>

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3 years ago
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