Answer:
1400 N
Explanation:
Verá, durante el salto mortal, el piloto se mueve en una trayectoria circular y la fuerza que actúa sobre él es una fuerza centrípeta.
Sea la fuerza centrípeta F, la masa del piloto (m) = 70 Kg, el radio (r) = 500 my la velocidad (v) = 360 km / hr * 1000/3600 = 100 m / s
F = mv ^ 2 / r
F = 70 * (100) ^ 2/500
F = 1400 N
A device used to initiate and control a sustained nuclear chain reaction.
<span>1.0x10^3 Joules
The kinetic energy a body has is expressed as the equation
E = 0.5 M V^2
where
E = Energy
M = Mass
V = Velocity
Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion
E = 0.5 * 7.2 kg * (17 m/s)^2
E = 3.6 kg * 289 m^2/s^2
E = 1040.4 kg*m^2/s^2
E = 1040.4 J
So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
The x- and y-coordinates are 9142.57 m and -304.425 m
<u>Explanation:</u>
As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell
in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of
and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by


For motion with constant acceleration, we know

Along the horizontal, x-axis, we might write this as

Measuring distances relative to the firing point means

we know that,

or,

By applying the values, we get,

The acceleration of gravity is vertically downward and is
, hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

we know,
and
, so,

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m