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Karo-lina-s [1.5K]

2 years ago

12

If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no

rth celestial pole?

1 answer:

kirill115 [55]2 years ago

8 0

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

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Which of the following is the name of the process scientists use to gain

harkovskaia [24]

Answer:

b

Explanation:

.

6 0

3 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed $v$ of the geosynchronous satellite is given by:

$v=\frac{2\pi R}{T}$

Because $2\pi R$ is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

$F_c=\frac{mv^{2} }{R}$

If we substitute the expression for $v$ in this formula, we get:

$F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}$

Since the centripetal force is the gravitational force $F_g$ between the satellite and the Earth, we know that:

$F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }$

Where G is the gravitational constant ( $G=6.67*10^{-11} Nm^{2}/kg^{2}$ ) and M is the mass of the Earth ( $M=5.97*10^{24}kg$ ). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

$R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km$

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0

3 years ago

What happens to electrical energy that is used by objects in our homes? (1 point) a It is absorbed by batteries. b It is destroy

Reil [10]

Question:

<em>What happens to electrical energy that is used by objects in our homes? (1 point)</em>

<em>a It is absorbed by batteries. </em>

<em>b It is destroyed. </em>

<em>c It is stored in solar panels. </em>

<em>d It is transformed into other forms of energy.</em>

<em />

Answer:

D

3 0

2 years ago

Read 2 more answers

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

2 years ago

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0

2 years ago