Complete Question:
Football player A has a mass of 110 kg, and he is running down the field with a velocity of 2 m/s. Football player B has a mass of 120 kg and is stationary. What is the total momentum after the collision?
Answer:
Total momentum = 220 Kgm/s.
Explanation:
<u>Given the following data;</u>
For footballer A
Mass, M1 = 110kg
Velocity, V1 = 2m/s
For footballer B
Mass, M1 = 120kg
Velocity, V1 = 0m/s since he's stationary.
To find the total momentum;
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
a. To find the momentum of A;

Momentum A = 220 Kgm/s.
b. To find the momentum of B;

Momentum B = 0 Kgm/s.
c. To find the total momentum of the two persons;
Substituting into the equation, we have;

<em>Total momentum = 220 Kgm/s. </em>
Are there any options??
I would have to say metal of course but without options I can't assume anything
Red shift of distant galaxies
Answer:
W = -0.480 J
Explanation:
given,
q₁ = 4 μC
q₂ = -4.10 μC


b = 0.381
k = 8.99 × 10⁹ Nm²/C²

![W = [-147.436\times (5.88-2.62)\times 10^{-3}]J](https://tex.z-dn.net/?f=W%20%3D%20%5B-147.436%5Ctimes%20%285.88-2.62%29%5Ctimes%2010%5E%7B-3%7D%5DJ)
W = -0.480 J
Work done by the electric force W = -0.480 J
Speed = (distance covered) / (time to cover the distance)
= ( 8.45 km) / (0.65 hr)
= (8.45 / 0.65) km/hr
= 13 km/hr