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.
The best and most correct answer among the choices provided by the question is <span>B.sound waves</span><span>.
</span>
<span>Particles move together or apart parallel to the direction of the sound wave.
</span>
Hope my answer would be a great help for you.
If you have more questions feel free to ask here at Brainly.
Answer:
Note that there is little variation among the transition metals. Electronegativities generally decrease from top to bottom within a group due to the larger atomic size. Of the main group elements, fluorine has the highest electronegativity (EN = 4.0) and cesium the lowest (EN = 0.79).
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval,

:

First we need to find the net charge flowing at a certain point of the wire in one second,

. Using I=0.92 A and re-arranging the previous equation, we find

Now we know that each electron carries a charge of

, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
Answer:
Daylight hours would be shorter.
Explanation:
If there were no tilt of the axis at all, every place on the planet except the north and south poles would have 12 hours of daylight and 12 hours of night every day of the year.
At a 10° tilt, the arctic circle and antarctic circles would be a little less than half the distance from the poles as they are today.