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Sonbull [250]
3 years ago
9

Which of the following terms describes the primary core of a comet?

Physics
1 answer:
Naddik [55]3 years ago
8 0
I would say that the nucleus is the best choice because it is the main solid part of the comet. Hope This Helped!
You might be interested in
If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?
mamaluj [8]

Answer:

Explanation:

KE=\frac{1}{2}mv^2

784=\frac{1}{2}(40)v^2

784=20v^2

39.2=v^2

v=6.26m/s

4 0
3 years ago
Pascal's Principle states that (a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished
nika2105 [10]

Answer:

a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished at every point in the fluid and on the walls of the container.

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area(A2) of the piston.

P=F/A

P1=P2

F1/ A1= F2/ A2

F2= F1* A2/ A1

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In an incompressible liquid, the volume and amount of mass does not vary when pressure is applied.

5 0
3 years ago
The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = \frac{1}{3} g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = \frac{1}{3}  m * g₁ = \frac{1}{3} w₁

⇒Weight of probe on Mars =\frac{1}{3} Weight of probe on earth

6 0
3 years ago
Which of the following provides evidence that geologic activity occurred in the past on Earth's Moon?
Bingel [31]

Answer: d) the presence of solidified lava flows on the Moon

Explanation:

A geological activity means an occurrence of event such as volcanic eruption, earthquake, sedimentation, erosion etc. The revolution of the Moon around the Earth , the axial tilt of the Moon  or the phases of the Moon are not surface features. hence, these events cannot provide the evidence of geological activity in the past of Moon.

The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.

6 0
3 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
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