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svetlana [45]
3 years ago
12

Use complete sentences to describe an extra-solar (exo) planet

Physics
2 answers:
Jobisdone [24]3 years ago
5 0
An exo planet is a planet that orbits a stat outside of the solar system.
anyanavicka [17]3 years ago
4 0

The correct answer of this question : The planet outside of our solar system.

EXPLANATION:

The extra-solar planet is the planet which is present beyond our solar system. It is also known as exoplanet.

Just like our solar system planets like the earth,Mars, Jupiter etc, the exoplanet is also a part of another solar system. The planets revolves around their sun i.e other star in different orbits. This star is the centre of their solar system.

Hence, the exact explanation for the exoplanet is the planet situated beyond our solar system, and revolves around any other centrally situated star.


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An experiment is designed to investigate the relation between the pressure and temperature in a tank that has a constant volume
viva [34]

Answer:

3rd order polynomial

Explanation:

Given that the increase in the order of the polynomial the error between the curve fit and measured data will decreases hence  :

The polynomial order that is best to use is the 3rd order polynomial, this is because using a 3rd order polynomial will produce a less variance and a low Bias

4 0
2 years ago
If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :
gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

  • c=1

So

  • b-3c=1
  • b-3=1
  • b=1+3
  • b=4

and

  • -a-b=-1
  • -a-4=-1
  • -a=-1+4=3
  • a=-3

So the unit is

  • DV⁴/F³
5 0
2 years ago
Read 2 more answers
What electrons are in “motion”, what do you have?
Sergeeva-Olga [200]

Answer:

When a positive charged object is placed near a conductor electrons are attracted the the object. ... When electric voltage is applied, an electric field within the metal triggers the movement of the electrons, making them shift from one end to another end of the conductor. Electrons will move toward the positive side. As you know, electrons are always moving. They spin very quickly around the nucleus of an atom. As the electrons zip around, they can move in any direction, as long as they stay in their shell.

7 0
3 years ago
26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the
denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

U=mg \Delta h

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h = 4 m + 0.02 m = 4.02 m is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J

4 0
2 years ago
A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0
2 years ago
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