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erastova [34]
10 months 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]10 months 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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Mation about your sour
Ann [662]

(#1). (D).

(#2). (C).

8 0
3 years ago
How does a generator produce an electric current
love history [14]

Electric generator is a device that converts mechanical energy obtained from an external source into electrical energy as an output. It was discovered that the above flow of electric charges could be induced by moving an electrical conductor such as a wire that contains electric charges in a magnetic field

4 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov
Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0
3 years ago
A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter
expeople1 [14]

Answer: 56.87m/s^{2}

Explanation:

If we make an analysis of the net force F_{net} of the rock that was thrown upwards, we will have the following:

F_{net}=F_{up}-W  (1)

Where:

F_{up}=200N is the force with which the rock was thrown

W is the weight of the rock

Being the weight the relation between the mass m=3kg of the rock and the acceleration due gravity g=9.79m/s^{2} :

W=m.g=(3kg)(9.79m/s^{2}) (2)

W=29.37 N (3)

Substituting (3) in (1):

F_{net}=200N-29.37 N  (4)

F_{net}=170.63 N  (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

F_{net}=m.a  (6)

Finding the acceleration a:

a=\frac{F_{net}}{m}  (7)

a=\frac{170.63 N}{3kg} (8)

Finally:

a=56.87m/s^{2}

3 0
3 years ago
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