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ipn [44]
3 years ago
10

Question 5 (2 points)

Physics
1 answer:
Radda [10]3 years ago
5 0

Answer:

Increase the work being done or decrease the time in which the work is completed

Explanation:

I got it right on the quiz i just took :)

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Que es la friccion <br> Cual es la primera ley de newton
lidiya [134]

Answer:

huh?...................

5 0
3 years ago
A light ray is striking the surface of Earth. Which factor would make the light ray more likely to be absorbed than reflected? l
Vladimir [108]

Answer:

B

Explanation:

Trust

5 0
3 years ago
Read 2 more answers
The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be
Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

#SPJ4

   

6 0
2 years ago
A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the
mash [69]

Answer:

\Delta x = v_0 t + \frac{1}{2}at^2

Explanation:

To solve this problem, we can use the following suvat equation:

\Delta x = v_0 t + \frac{1}{2}at^2

where

\Delta x is the vertical displacement of the frog

v_0 is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from v_0, all the other quantities are known. In fact:

\Delta x =0.1 m is the vertical displacement

t = 2 s is the total time of flight

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

Therefore, solving for v_0, we find the initial velocity of the frog:

v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s

4 0
3 years ago
Which of the following is not a reason that coral reefs are important?
Brut [27]
My Answer: They contain a high diversity of organisms.

Hope I helped! :D
3 0
3 years ago
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