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Norma-Jean [14]
3 years ago
11

From the earth, the moon subtends an angle of approximately 0.5°. If the distance to the moon is approximately 240,000 miles, fi

nd an approximation for the diameter of the moon accurate to the nearest hundred miles.
Physics
2 answers:
slega [8]3 years ago
8 0

The approximation for the diameter of the moon accurate to the nearest hundred miles is s =  +- 2100 miles

<h3>Further explanation </h3>

Arc Measure definition is,  in a circle, arc degree measure is equal to the measure of the central angle that intercepts the arc. Whereas the arc Length definition is, In a circle, the length of an arc is a portion of the circumference. Where the letter "s" is used to represent arc length.      

s = r * \theta (Radian Measure and Arc of a Circle).

There is a formula which relates the arc length of a circle of radius, r, to the central angle, θ in radians.

The radian measure θ of a central angle is defined as the ratio of the length of the arc  the angle subtends s, divided by the radius of the circle r.

where s = the arc length

\theta = angle  in radians

0.5 degrees *  \frac{\pi}{180}  = 0.008726646 radians \\ r = 240000 miles  \\ s  = 240000 * 0.008726646 \\

s = 2094.4 miles or

s =  +- 2100 miles

So the approximation for the diameter of the moon accurate to the nearest hundred miles is s =  +- 2100 miles

<h3>Learn more</h3>
  1. Learn more about the radian measure of the angle brainly.com/question/4675968

<h3>Answer details</h3>

Grade:  9

Subject:  physics

Chapter:  the radian measure of the angle

Keywords:  moon

laiz [17]3 years ago
3 0
The subtended angle and by an arc is given by:
S = r∅
Although the arc is circular, the huge amount of distance means that it is almost equivalent to the director.
D = 240,000 x (0.5 x π)/180
D = 2094 miles
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natulia [17]

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Explanation:

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Here L is the lenght of the wire, A the area and p is the resistivity of wire.

As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.

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6 0
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Read 2 more answers
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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Ilia_Sergeevich [38]
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(I will explain the second question with an example, so it's easier to understand)
-For Sideral month for example of the moon it cactually complete one revolution in around 27.3 days.
However, since the earth moves, for us it took some more time to see the moon the same as before (fullmoon to fullmoon) again. That make synodic month of the moon to be around 29.5 days.
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