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3 years ago

What are some of the challenges that you might face if you wanted to observe and study stars from Earth?

2 answers:

7 0

<span>The earth moves, which is unstoppable and if the earth moves the telescope won't be able to see it clearly because the telescope needs to be able to move at the same pace as earth to keep up to the objects</span>

AlexFokin [52]3 years ago

4 0

Another main issue with astronomical sighting is light pollution. The lights from our cities and such drown out the light of the stars. You could also say clouds, humidity, fog, or anything of the kind.

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A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

3 years ago

If you are standing at a beach on Earth at the same time that the shadow of the moon falls across your location, what event woul

Tcecarenko [31]

A solar eclipse .....................

5 0

3 years ago

A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p

Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d = 2.0 mm = 2 x 10⁻³ m

magnitude of the electric field = 200 kN/C

Capacitance of the capacitor is calculated as follows;

$C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F$

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

4 0

3 years ago

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

Drupady [299]

Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series & also Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

= 20 ║ 15

= (20×15) / (20 + 15)

= 8.57 ohms

Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

5 0

3 years ago

What is the weight of a person with a massa of 80kg

Archy [21]

784 Newtons or 176.37 lbs

5 0

3 years ago