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3 years ago

Give at least two reasons today’s astronomers are so interested in the discovery of additional Earth-approaching asteroids.

Physics

1 answer:

blondinia [14]3 years ago

4 0

Answer:

1. Potential hazard

2. Mining opportunity

Explanation:

The two reason, why astronomers are so interested in the discovery of additional Earth-approaching asteroids:

1. Potential hazard: We have proof that the dinosaurs got extinct because of an asteroid/comet strike on Earth. Also we have seen the effects of the Tunguska event and Chelyabinsk tragedy. These are enough to show us that asteroids can be very dangerous and wipe out the life from Earth.

2. Mining Opportunity: We have discovered a lot of asteroids which contains a lot of metal and precious elements. There can be a possibility of mining such asteroids in the future and reducing the burden on Earth.

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All else equal, the payback period for a project will decrease whenever the_______.

telo118 [61]

Answer:

(B) cash inflows are moved earlier in time.

Explanation:

The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form

The formula to compute the payback period is shown below:

Payback period = Initial investment ÷ Net cash flow

where,

The net cash flow = annual net operating income + depreciation expenses

The payback period of the project decreases when the accumulated starting year cash flows increases that results the movement of the cash inflows earlier in time

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2 years ago

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The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.

iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

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2 years ago

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a _______ wave is a mechanical wave in which particles of the medium vibrate in a direction that is parallel to the direction in

Helga [31]

The answer is Longitudinal

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2 years ago

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Which of the following is fact-based science rather than part of a personal belief system?

marusya05 [52]

Since there are no choices, then this question calls for open-ended answers. Facts-based science must have proven underlying laws that support inferences such as Coulomb's Law, Kinetic Theory of Matter and many more. On the other hand, examples of science that focus on personal belief is philosophy. This depends on the perspective of known philosophers. An example would be Sigmund Freud who proposed the theory of 3 personalities. Although it is more on personal beliefs, this is used as a foundation in the study of psychology.

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2 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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2 years ago