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kow [346]
3 years ago
8

two soccer players run toward each other. one player has a mass of 85 kg and runs west with a speed of 8 m/s, while the other ha

s a mass of 105 kg and runs east with a speed of 7 m/s. what is the total momentum of the system made up of the two players?
Physics
2 answers:
marissa [1.9K]3 years ago
8 0

Answer:

Total momentum, p = 55 kg-m/s

Explanation:

It is given that,

Mass of player 1, m₁ = 85 kg

Mass of player 2, m₂ = 105 kg

Speed of player 1, v₁ = -8 m/s (west)

Speed of player 2, v₂ = 7 m/s (east)

Momentum is equal to the product of mass and velocity. For this system, momentum is given by :

p=m_1v_1+m_2v_2

p=85\ kg\times (-8\ m/s)+105\ kg\times 7\ m/s

p = 55 kg-m/s

The total momentum of the system made up of the two players is 55 kg-m/s.

adelina 88 [10]3 years ago
6 0

Answer:

+55 kg m/s  

Explanation:

Momentum is the product mass and velocity.

P = m × v

Let the east be in positive direction and west be in negative.

mass of player 1, m = 85 kg

velocity of player 1, v = -8 m/s

mass of player 2, m' = 105 kg

velocity of player 2, v' = +7 m/s

In the absence of any external unbalanced force acting on the system, the total momentum remains conserved. So, the total momentum of this system made having two players is:

P = m v + m' v'

⇒P = (85)(-8) + (105)(7)

⇒P = -680 + 735 = +55 kg m/s

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Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou
SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

7 0
4 years ago
Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm
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Answer:

Explanation:

Potential energy of the system of charges

=  9 x 10⁹ x [  q₁q₂ / r₁₂ +  q₂q₃ / r₂₃ +  q₁q₃ / r₁₃ ]

here  r₁₂ ,  r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy  =  9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08  +  -200x10⁻¹⁸ / .04 +  200 x 10¹⁸ / .04 ]

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b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

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5 0
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