Answer:
(a) The speed of the target proton after the collision is:
, and (b) the speed of the projectile proton after the collision is:
.
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:
, and y axle:
. Now replacing the value given as:
,
for the projectile proton and according to the problem
are perpendicular so
, and assuming that
, we get for x axle:
and y axle:
, then solving for
, we get:
and replacing at the first equation we get:
, now solving for
, we can find the speed of the projectile proton after the collision as:
and
, that is the speed of the target proton after the collision.
Halite or sulfur or gold or silver
Answer:
Vagetagble, fruits,fresh milk and juice vitamins c
Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa