Physics, 10.03.2021 16:10 queenbrebre2294

football player A has a mass of 110 kg is running on the field is velocity of 2 m/s football player B has a mass of 120 kg and a stationary what if football player a collage of the football player babe what is the total momentum after the collision

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Answer from: JANA279

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?

Total momentum = 220 Kgm/s.

Explanation:

Given the following data;

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;

Momentum = mass * velocity

a. To find the momentum of A;

$Momentum \; A = 110 * 2$

Momentum A = 220 Kgm/s.

b. To find the momentum of B;

$Momentum \; B = 120 * 0$

Momentum B = 0 Kgm/s.

c. To find the total momentum of the two persons;

$Total \; momentum = Momentum \; A + Momentum \; B$

Substituting into the equation, we have;

$Total \; momentum = 220 + 0$

Total momentum = 220 Kgm/s.

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

Answer from: Quest

force required to detach it is given by

$f = 4 * 10^{-8} n$

charges are taken to be fundamental charge

$q_1 = q_2 = 1.6 * 10^{-19}c$

now we have

$f = \frac{kq_1q_2}{r^2}$

$4 * 10^{-8} = \frac{9*10^9 * 1.6*10^{-19}* 1.6* 10^{-19}}{r^2}$

$r = 7.6 * 10^{-11} m$

so above is the minimum distance at which it will detach

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