C. The downward component of the projectile's velocity continually increases
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion (with constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction
Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:
where
u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)
downward is the acceleration of gravity
t is the time
So the equation becomes
This means that
C. The downward component of the projectile's velocity continually increases
Because every second, it increases by 
in the downward direction.
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Answer:
subs
Explanation:
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Answer: a) Mr = 2.4×10^-4kg/s
V = 34.42m/a
b) E = 173J
Ø = 2693.1J
c) Er = 0.64J/s
Explanation: Please find the attached file for the solution
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
