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Alla [95]
3 years ago
7

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

00 km. Use scientific notation to find the number of kilometers in one orbit. Write your answer to two decimail places and don't forget the units.
Physics
1 answer:
madreJ [45]3 years ago
8 0

Answer:

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km

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A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0
3 years ago
Which property of light would provide evidence for the idea that light is a wave?
Lena [83]

Color property of light would provide evidence for the idea that light is a wave

<h3><u>Explanation:</u></h3>

The reality is that light manifests practices that are representative of both waves and particles. Young proposed that light of varying colors was formed of waves possessing various lengths, a basic theory that is popularly believed today. In contradiction, the particle theory advocates envisioned that several colors were obtained from particles holding either various masses or moving at various speeds.

All waves are perceived to experience refraction when they transpire from one means to another means. Light, similar to any wave, is apprehended to refract as it transfers from one medium into another medium.

5 0
3 years ago
How much force is required to accelerate 3 kg at 3 m/sec^2
bogdanovich [222]

Answer:

F=?

a=3m/s2

m=3kg

F=ma=[3][3]=9 N

Explanation:

7 0
3 years ago
Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. An
Vikentia [17]

Answer: Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Another bright star, Regulus, has a parallax of 0.042 arcseconds. Then, the distance in parsecs will be,23.46.

Explanation: To find the answer, we have to know more about the relation between the distance in parsecs and the parallax.

<h3>What is the relation between the distance in parsecs and the parallax?</h3>
  • Let's consider a star in the sky, is d parsec distance from the earth, and which has some parallax of P amount.
  • Then, the equation connecting parallax and the distance in parsec can be written as,

                                     d=\frac{1}{P}

  • We can say that,

                                    dP=constant.\\thus,\\d_1P_1=d_2P_2

<h3>How to solve the problem?</h3>
  • We have given that,

                                     d_1=2.6 parsecs.\\P_1=0.379arcseconds.\\P_2=0.042 arcseconds.\\d_2=?

  • Thus, we can find the distance in parsecs as,

                                     d_2=\frac{d_1P_1}{P_2} =23.46 parsecs

Thus, we can conclude that, the distance in parsecs will be, 23.46.

Learn more about the relation connecting distance in parsecs and the parallax here: brainly.com/question/28044776

#SPJ4

6 0
2 years ago
Rank in order, from smallest to largest, the six torques (torque 1 to torque 2) about the hinge.?
anzhelika [568]
The solution for this problem is:torque1 = torque2 = FL / 2 
Torque 3 = Torque 4 = FL / 2 * sin (theta) 
Torque 5 = 2 FL 
Torque 6 = 0 

So the order of the torques from smallest to largest is torque 6, (torque 3 and 4), (torque 1 and 2), torque 5.


Remember or just take note that sin (theta) < 1 is why 3 and 4 are less than 1 and 2.
5 0
3 years ago
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