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erastova [34]
1 year ago
13

a new planet is found with a density one third as much at earth and a radius twice that of earth. what is the acceleration due t

o gravity on this new planet?
Physics
1 answer:
IrinaVladis [17]1 year ago
8 0

The acceleration due to gravity on the new planet would be one third that of Earth's, or about 9.8 m/s^2.

What is planet?
This question might seem to have an easy solution, but it doesn't. The planets Earth, Mars, and Jupiter are well known. But before recent discoveries sparked a passionate debate about how to best characterise them, both Pluto and Ceres have been thought of as planets. This debate is still going strong today. The International Astronomical Union adopted the most recent definition of the a planet in 2006.

It specifies three requirements for a planet:

-It must move about a star (in our cosmic neighborhood, the Sun).
-It must be large enough for gravity to pull it in a spherical direction.
-It must have been large enough that any nearby objects of the a similar size were removed by its gravitational pull.

Given that the mass of the new planet is one third that of Earth's and its radius is twice that of Earth's, we can say that its volume is eight times that of Earth's.  This means that its average density is one third that of Earth. Therefore, the acceleration due to gravity on the new planet would be one third that of Earth's, or about 9.8 m/s^2.

To learn more about planet
brainly.com/question/13106554
#SPJ4

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A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was
Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

7 0
3 years ago
Help, please. I am not sure what to do.
miss Akunina [59]

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

3 0
2 years ago
Yoyoyoyoyoyoyoyoyoyoyoyoyoyoyoy
Vikentia [17]

Answer:

Yoyoyoyoyooyoyy

Explanation:

Yoyoyoyoyoyoyoyoyooy

4 0
3 years ago
Read 2 more answers
I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

7 0
3 years ago
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
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