I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at 
Solving for the angle, we get
or
![\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%20%20%5Csin%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7B34%5C%3A%5Ctext%7BN%7D%7D%7B%285.1%5C%3A%5Ctext%7Bkg%7D%29%289.8%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%5Cright%5D)
Answer:
Explanation:
kinetic energy required = 1.80 MeV
= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 2.88 x 10⁻¹³ J
If v be the velocity of proton
1/2 x mass of proton x v² = 2.88 x 10⁻¹³
= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³
v² = 3.45 x 10¹⁴
v = 1.86 x 10⁷ m /s
If V be the potential difference required
V x e = kinetic energy . where e is charge on proton .
V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³
V = 1.8 x 10⁶ volt .
