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Gemiola [76]
3 years ago
13

Consider again the objects you ranked by distance. Suppose each object emitted a burst of light right now. Rank the objects from

left to right based on the amount of time it would take this light to reach Earth, from longest time to shortest time.a. star on far side of Andromeda Galaxyb. star on near side of Andromeda Galaxyc. star on far side of Milky Way Galaxyd. star near center of Milky Way Galaxy1. Orion Nebula2. Alpha Centauri3. Pluto4. The Sun
Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

Following are the solution to this question:

Explanation:

That light takes a very long time to hit the planet, and the object is far off the earth. The light of such an item near to the planet takes less time to enter it. The star is 2,5 million light-years from the Planet on the far side of the Andromeda Galaxy. But on the other hand, the moon is 15 crore miles from the earth, so sunlight is quickly reached on the ground as the other thing.  

That milky way away from the earth is 66,500 light-years far, that distance between Earth and Orion nebula is 1,344 light-years, with such a distance of 4,367 light-years. The earth is 5.2261 trillion km apart from Pluto.

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What happens to a circuit's resistance (R), voltage (V), and current (1) when
hammer [34]

Answer:

D.

R increases

V is constant

I decreases

Explanation:

The resistance of a wire is given by the following formula:

R = \frac{(Resistivity)(L)}{A}

It is clear from this formula that resistance is directly proportional to the length of wire. So, when length of wire is increased, <u>the resistance of circuit increases</u>.

The <u>voltage in the circuit will be constant</u> as the voltage source remains same and it is not changed.

Now, we can use Ohm Law:

V = IR

at constant V:

I ∝ 1/R

it means that current is inversely proportional to resistance. Hence, the increase of resistance causes <u>the current in circuit to decrease.</u>

Therefore, the correct option will be:

<u>D.</u>

<u>R increases </u>

<u>V is constant </u>

<u>I decreases</u>

6 0
2 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
một hòn đá ném theo phương nằm ngang với vân tốc v0=15 m/s. tính gia tốc tiếp tuyến và gia tốc pháp tuyến của hòn đá sau lúc ném
Lady_Fox [76]

Answer:

what language is this???

5 0
3 years ago
The ___ energy in a mechanical system is determined by adding the potential and kinetic enters together
Svetllana [295]

Answer:

A, total.

<em>The </em><em>total</em><em> energy in a mechanical system is determined by adding the potential and kinetic enters together.</em>

<em />

<u><em>i hope this helped at all.</em></u>

<em />

4 0
3 years ago
The proper IUPAC naming convention for the compound symbol 'KI' is
Darya [45]
Proablyly a or c idk 
6 0
3 years ago
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