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alukav5142 [94]
3 years ago
13

From the center of the Earth to the moon, what should the orbital radius of such satellite be in order to stay over the same poi

nt on the earth’s surface?​
Physics
1 answer:
yulyashka [42]3 years ago
3 0

In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.

But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.

The Moon has nothing to do with any of this.

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3. What will happen to the black bass and blue gill as the floor of the ponds fills with organic
yKpoI14uk [10]

Answer: die

Explanation: oyxagan all goon bc of all dat suffs

8 0
2 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
Why is this statement false?
koban [17]

The statement is false because as the temperature of water is decreased from 4C to 0C, <em>expansion</em> occurs.

You should understand that this is very weird behavior for any compound ... within a small temperature range, the stuff actually gets bigger as it gets colder.  That's why cubes float in your sody, and bergs float in the ocean.

Water is one of only two known substances that do this.  And if water didn't do it, then life on Earth would not be possible.  

Hmmm....

7 0
2 years ago
In 8.5 s a fisherman winds 2.4 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)
SSSSS [86.1K]

Answer:

9.412 rad/s.

Explanation:

Velocity is the rate of change of an object's position.

V = x/t

Where x is the distance in m

= 2.4 m

t is time taken in s

= 8.5 s

V = 2.4/8.5

= 0.2824 m/s.

Equating linear velocity and angular velocity,

V = ω*r

Where,

ω Is the angular speed in rad/s

r is the radius of the circle in m

= 3 cm

= 3cm * 1m/100 cm = 0.03 m

ω = V/r

= 0.2824/0.03

= 9.412 rad/s.

4 0
3 years ago
You are given two temperatures for substance: one is the melting point and one is the boiling point. How do you determine which
Sergio039 [100]

melting point is where the substance is melting and turning from a solid to a liquid.

boiling point is where molecules vibrate and produce energy which makes the water hot and becomes a boil.

8 0
3 years ago
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