What does the abbreviation DC stand for in physics? A.direct charge B.dark circuit C.direct current D. drop charge
1 answer:
You might be interested in
Let's solve the problem step-by-step.
A) Let's start writing the laws of motion on both x (horizontal) and y (vertical) axis. On the x-axis, it's a uniform motion with constant velocity, while on the y-axis it's an uniformly accelerated motion:
where h=654 m is the initial height of the projectile,
and 
are the initial velocities on the x- and y-axis, g=9.81 m/s^2 and t is the time. ad 
have a negative signs because they both point downward.
We can calculate the initial velocity on the y-axis by requiring that
, since we know that after 6.70 s the projectile reached the ground. Therefore:
from which we find
Then we can find the magnitude of the initial velocity v using the angle with respect to the vertical
:
B) Let's calculate the component of the initial velocity on the x-axis:
And then, we can find how far the projectile traveled horizontally by calculating Sx at t=6.70 s, when it hits the ground:
C) The horizontal component of the velocity does not change during the motion, since it's an uniform motion on the x-axis. Therefore, vx at t=6.7 s is the same as its initial value:
D) Instead, vy changes during the motion since it's an accelerated motion, following the law
Using t=6.7 s, we can find the vertical velocity just before the projectile hits the ground
Writing it with positive sign:
A) Let's start writing the laws of motion on both x (horizontal) and y (vertical) axis. On the x-axis, it's a uniform motion with constant velocity, while on the y-axis it's an uniformly accelerated motion:
where h=654 m is the initial height of the projectile,
We can calculate the initial velocity on the y-axis by requiring that
from which we find
Then we can find the magnitude of the initial velocity v using the angle with respect to the vertical
B) Let's calculate the component of the initial velocity on the x-axis:
And then, we can find how far the projectile traveled horizontally by calculating Sx at t=6.70 s, when it hits the ground:
C) The horizontal component of the velocity does not change during the motion, since it's an uniform motion on the x-axis. Therefore, vx at t=6.7 s is the same as its initial value:
D) Instead, vy changes during the motion since it's an accelerated motion, following the law
Using t=6.7 s, we can find the vertical velocity just before the projectile hits the ground
Writing it with positive sign: