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3 years ago

_____ helped to socially and politically unify the early byzantine empire by tolerating and even promoting christianity.

2 answers:

Ber [7]3 years ago

8 0

Constantine I helped to socially and politically unify the early byzantine empire by tolerating and even promoting christianity.

Fittoniya [83]3 years ago

8 0

Answer:

Constantine I

Explanation:

APEX

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How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.

goldfiish [28.3K]

143m/s if you just perhaps by what you know you'll figure it out

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3 years ago

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

lbvjy [14]

Answer:

The beam of light is moving at the peed of:

$\frac{dy}{dt} = \frac{80\pi}{3}$ km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, $\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi$ (1)

Now, in the right angle in the given fig.:

$tan\theta' = \frac{y}{3}$

Now, differentiating both the sides w.r.t t:

$\frac{dtan\theta'}{dt} = \frac{dy}{3dt}$

Applying chain rule:

$\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}$

$sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}$

Now, using $tan\theta = \frac{1}{m}$ and y = 1 in the above eqn, we get:

$(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}$

Also, using eqn (1),

$8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}$

$\frac{dy}{dt} = \frac{80\pi}{3}$

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3 years ago

A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d

OleMash [197]

Answer:

W = 0.060 J

v_2 = 0.18 m/s

Explanation:

solution:

for the spring:

W = 1/2*k*x_1^2 - 1/2*k*x_2^2

x_1 = -0.025 m and x_2 = 0

W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

W = 0.060 J

the work-energy theorem,

W_tot = K_2 - K_1 = ΔK

with K = 1/2*m*v^2

v_2 = √2*W/m

v_2 = 0.18 m/s

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3 years ago

Each inner energy level of an atom has a maximum number of it can hold

CaHeK987 [17]

That would be a maximum of 4 atoms

3 0

3 years ago

Read 2 more answers

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago