Answer:
vx = 65 yd/3 sec = 21.7 yd/sec since horizontal speed is constant
vy = g t = (32 / 3) yd/sec^2 * 1.5 sec = 16 yd/sec where 32/3 is the acceleration due to gravity in yds / sec^2 and 1.5 is the time to travel each way in the vertical direction
V = (vx^2 + vy^2)^1/2 = (21.7^2 + 16^2)^1/2 = 27 yd/sec
tan theta = vy/vx = 16 / 21.7 = .737 theta = 36.4 deg
You can check using the range formula:
R = v^2 sin (2 theta) / g = 27^2 * .955 / (32 / 3) = 65.3 yds
The difference from 65 yds may be rounding error.
The electron is accelerated through a potential difference of

, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:

where
m is the electron mass
v is the final speed of the electron
e is the electron charge

is the potential difference
Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:

where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
Answer:
Acceleration, a = -31.29 m/s²
Explanation:
It is given that, Initial speed of the aircraft, u = 140 mi/h =
62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
a = t
0-62.58 2 a =
a = -31.29 m/s²
So, the acceleration of the aircraft is 31.29 m/s²
and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)