To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by
Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So
x = Displacement
t = time
With the data we have we can find the time as well
With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,
Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
The potential energy of an object is defined by the equation: PE = mgh, where m = the mass of the object, g = the gravitational acceleration and h = the object's height above the ground.
Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.
Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
Its acceleration is constant.
