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AnnZ [28]
2 years ago
15

The solar mass of the Sun is 1. The orbital period of Jupiter is 11. 9 Earth years. What is the distance between Jupiter and the

Sun? 5. 2 AU 41 AU 52 AU 410 AU.
Physics
1 answer:
cestrela7 [59]2 years ago
6 0

The distance between Jupiter and the sun is 5.2 AU.

According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.

r =T^2/3

r = (11.9)^2/3

r = 5.2 AU

Learn more: brainly.com/question/15207516

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A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn
vesna_86 [32]

The correct answer of this question is :  A) Change alternating current into direct current.

EXPLANATION :

As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .

The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.

Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.


4 0
3 years ago
Read 2 more answers
34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf
ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

d_1 = \frac{1}{2}(0.6) t^2

also the distance moved by passenger

d_2 = \frac{1}{2}(1.2) t^2

so we will have

d_1 + 5 = d_2

0.3 t^2 + 5 = 0.6 t^2

0.3 t^2 = 5

t = 4.08 s

3 0
3 years ago
To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,
bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1  = 800 Hz

harmonic frequency f2  = 1120 Hz

harmonic frequency f3  = 1440 Hz

solution

first we get here fundamental frequency that  is express as

2F = f2 - f1    ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength  = Speed ÷ Fundamental frequency    ................2

here speed of waves in air  = 340 m/s

so put here value

Wavelength  =340 ÷ 160

Wavelength   = 2.125 m

so

Length of the pipe will be

Length of the pipe = 0.25 × wavelength    ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

7 0
3 years ago
Where did we use rotational and irrotational flow​
WARRIOR [948]

Answer:

The term rotational and irrotational flow is associated withe the flow of particles in fluid.

The common example of irrrotational flow can be seen on the carriages of the Ferris wheel (giant wheel).

Explanation:

  • If the fluid is rotating along its axis with the streamline flow of its particles,then this type of flow is rotational flow.
  • Similarly if fluid particles do not rotate along its axis while flowing in a stream line flow then it is considered as the irrotational flow.
  • In majority, if the flow of fluid is viscid then it is rotational.
  • Fluid in a rotating cylinder is an example of rotating flow.
3 0
3 years ago
A(n) 96.1 g ball is dropped from a height of 59.1 cm above a spring of negligible mass.The ball compresses the spring to a maxim
Serhud [2]

Answer:

Explanation:

Mass of ball Is m=96.1g=0.0961kg

Height above spring is 59.1cm

L=0.591m

Extension of the spring is 4.75403cm

e=0.0475403m

Then the distance the ball traveled is H=L+e

H=0.591+0.0475403

H=0.6385403m

Then, the potential energy of the ball is given as

P.E=mgh

P.E=0.0961×9.81×0.6385403

P.E=0.602J

From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another

Then, the P.E is transferred to the work done by the spring

Then, Work done by spring is given as

W=½ke²

W=P.E=½×k×0.0475403²

0.602=½×k×0.0475403²

k=0.602×2/0.0475403²

k=532.72N/m

The spring constant is 532.72 N/m

4 0
3 years ago
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