Answer:
Explanation:
On the Moon :----
1500 x 1.6 = 2400 m /s is initial velocity of bullet .
g = 1.6 m /s²
v = u - gt
0 = 2400 - 1.6 t
t = 1500 s
This is time of ascent
Time of decent will also be the same
Total time of flight = 2 x 1500 = 3000 s
On the Earth : ---
v = u - a₁ t
0 = u - a₁ x 18
u = 18a₁
v² = u² - 2 x a₁ x 2743.2
0 = (18a₁ )² - 2 x a₁ x 2743.2
a₁ = 16.93
For downward return
s = ut + 1/2 a₂ x t²
2743.2 = 0 + .5 x a₂ x 31²
a₂ = 5.7 m /s²
If d be the deceleration produced by air
g + d = 16.93 ( during upward journey )
g - d = 5.7
g = (16.93 + 5.7) / 2
= 11.315 m / s
d = 5.6 m /s²
So air is creating a deceleration of 5.6 m /s².
Answer:
Explanation:
From A to B
distance traveled with velocity 
in time 
from B to C
distance traveled is 0.5 d with 
and 
velocity for half-half time
divide 1 and 2 we get
Now average velocity is given by
taking common
This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:
Kinetic energy = (1/2) · (mass) · (speed²)
Momentum = (mass) · (speed)
So, now ... We know that
==> mass = 15 kg, and
==> kinetic energy = 30 Joules
Take those pieces of info and pluggum into the formula for kinetic energy:
Kinetic energy = (1/2) · (mass) · (speed²)
30 Joules = (1/2) · (15 kg) · (speed²)
60 Joules = (15 kg) · (speed²)
4 m²/s² = speed²
Speed = 2 m/s
THAT's all you need ! Now you can find momentum:
Momentum = (mass) · (speed)
Momentum = (15 kg) · (2 m/s)
<em>Momentum = 30 kg·m/s</em>
<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>
No it can't it's material
Answer:
D
Explanation:
appearance is not a imp factor . location could be imp. becoz a proper environment is need to study such courses.
