Answer:
<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>
Explanation:
The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:
• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>
• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.
• They emit sufficient radiation at wavelengths conducive to photosynthesis.
• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.
<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>
Force=tension-fg sin ∅
=140-mg sin 18.5
=140-124.35
=15.62N
a=f/m=15.62/40=0.39
now,
v²=u²+2as
=2×0.39×80
v²=62.4
v=7.8m/s
Answer: Option (a) is the correct answer.
Explanation:
Sunspots are defined as the dark and cooler areas on the surface of the Earth. A sunspot is present in a region known as photosphere.
Temperature of a sunspot is about 3,800 degrees Kelvin whereas photosphere has a temperature of about 5,800 degree kelvin.
Thus, we can conclude that out of the given options, the statement they are areas of the sun that are "burnt out" is false about sunspots.
Maybe nobody ever mentioned it to you, but it turns out that
current is another one of those things that's always conserved ...
it can't created or destroyed, just like energy and mass.
The total current in a circuit is always the same, but it can get
split up and travel through different paths for a while.
<span>==> The total current is just the amount of current
that's flowing in and out of </span><span>the battery.
Diagram #1).
</span>The total current coming out of the battery is 15 A.
That current is going to split up when it reaches the resistors.
Part of it will flow through each resistor, but both of them
will still add up to 15 A .
You have 9 A flowing through one resistor.
So the current in the other resistor is (15 - 9) =<span> 6 A.
Diagram #2).
</span>The total current coming out of the battery is 10 A.
That current is going to split up when it reaches the resistors.
Part of it will flow through each resistor, but all of them
will still add up to 10 A .
You have 2.5 A through one resistor and 3.5 A through another one.
So the amount left for the last resistor is (10 - 2.5 - 3.5) =<span> 4 A.</span>
Answer:
Explanation:
Given
charge on alpha particle=+2e
mass of alpha particle=
kg
Charge on gold nucleus=+79e
Velocity at r=1m is 
Using Energy conservation
Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus
therefore
![\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cleft%20%28%206.64%5Ctimes%2010%5E%7B-27%7D%5Cright%20%29%5Cleft%20%28%201.9%5Ctimes%2010%5E%7B7%7D%5E2%5Cright%20%29%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20158%5Ctimes%20%5Cleft%20%28%201.6%5Ctimes%2010%5E%7B-19%7D%5Cright%20%29%7D%7By%7D%5Cleft%20%5B%5Cfrac%7B1%7D%7Br_0%7D-%5Cfrac%7B1%7D%7B1%7D%5Cright%20%5D)
on solving we get
