Horizontal distance covered by a projectile is X = Vix *T
where Vix is the initial horizontal component of velocity and T is time taken by the projectile
Vix = ViCos theta
In question they said that initial velocity and angle is same on earth and moon
so Vix would remains same
now let's see about time taken T
time taken to reach the highest point
Vfy = Viy +gt
at highest point vertical velocity become zero so Vfy =0
0 = Vi Sin theta + gt
t = Vi Sintheta /g
Total time taken to land will be twice of that
On earth
Te= 2t
Te = 2Sinθ/g
on moon g is one-sixth of g(earth)
Tm = 2Sinθ/(g/6)
Tm = 6(2Sinθ/g)
Tm = 6Te
so total time taken by the projectile on moon will be six times the time taken on earth
From first equation X = Vix*T
we can see that X will also be 6 times on moon than earth
so projectile will cover 6 times distance on moon than on earth
Answer:
intinal speed should be 10
Explanation:
v = v0 + at
30 = v0 +10.2
then v0= 10
There are many porperties. You can use Altitude, Temperature, Pressure and Density, but the best one is temperature. The resaon for that is that based on the temperature changes then the athmosphere can be broken into four major layers. Remember that the layers are the following: <span>the </span>troposphere,the<span> </span>stratosphere, <span>the </span>mesosphere<span>, and the</span>thermosphere<span>.</span>
Answer:
The length of the tube is 3.92 m.
Explanation:
Given that,
Electric potential = 100 MV
Length = 4 m
Energy = 100 MeV
We need to calculate the value of
Using formula of relativistic energy
Put the value into the formula
Here,
We need to calculate the length
Using formula of length
Put the value into the formula
Hence, The length of the tube is 3.92 m.
<em>Energy</em><em> </em><em>can</em><em> </em><em>neither </em><em>be</em><em> </em><em>created </em><em>nor</em><em> </em><em>be</em><em> </em><em>destroyed</em><em> </em><em>but</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>converted</em><em> </em><em>from</em><em> </em><em>one</em><em> </em><em>form</em><em> </em><em>to</em><em> </em><em>another </em><em>.</em>