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SVEN [57.7K]
3 years ago
5

Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

st. What is the total momentum of the system
Physics
1 answer:
sukhopar [10]3 years ago
6 0

The total momentum of the system is equal to 50 Kgm/s.

<u>Given the following data:</u>

  • Mass 1 = 80 kg
  • Velocity 1 = 2 m/s east.
  • Mass 2 = 70 kg
  • Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

Momentum = mass \times velocity

<u>For Football player 1:</u>

Momentum = 80 \times 2

Momentum 1 = 160 Kgm/s.

<u>For Football player 2:</u>

Momentum = 70 \times 3

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160

Total momentum = 50 Kgm/s.

<u>Note:</u> We subtracted because the football players were moving in opposite directions.

Read more: brainly.com/question/15517471

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iren [92.7K]

Answer:

(a)  a₁:  jogger  acceleration= 1.5 m/s²

(b)  a₂:  car  acceleration = 1.5 m/s²

(b)  d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁:  jogger displacement   

t₁ :  jogger time

v₀₁:  jogger initial speed

vf₁:  jogger  final speed

a₁:  jogger  acceleration

d₂: car displacement   

t₂ : car  time

v₀₂: car  initial speed

vf₂:  car  final speed

a₂:  car  acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

(c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

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The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a
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Answer:

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Explanation:

We are given that

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