F = q₁q₂C / r²
F force
q charge
C Coulomb constant
r separation between charges
The photon can be absorbed and the energy of the photon is exactly equal to the energy-level difference between the ground state and the level d.
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback
So for conservation of momentum,
rho = mv
M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf
For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2
Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
Answer:
d = 10 inch
Explanation:
The farthest distance between the centers, is along the diagonal of the rectangle. Therefore, we need to calculate the diagonal of the rectangle, but counting the fact that we have both circles.
So if, one side is 12 inch, and the other is 14 inch, we can use the Pitagoras theorem which is:
d = √(a²) + (b)²
Where a and b, are the lenght of the rectangle, but without the lenght of the diameter of both circles.
With this, the expression is this:
d = √(14 - 6)² + (12 - 6)²
d = √64+36
d = √100
d = 10 inches
Answer: 909 m/s
Explanation:
Given
Mass of the bullet, m1 = 0.05 kg
Mass of the wooden block, m2 = 5 kg
Final velocities of the block and bullet, v = 9 m/s
Initial velocity of the bullet v1 = ? m/s
From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve
m1v1 = (m1 + m2) v, on substituting, we have
0.05 * v1 = (0.05 + 5) * 9
0.05 * v1 = 5.05 * 9
0.05 * v1 = 45.45
v1 = 45.45 / 0.05
v1 = 909 m/s
Thus, the original velocity of the bullet was 909 m/s
