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choli [55]
2 years ago
6

Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass

of 190 pounds and is running in the same direction at 3.0 mi/hr. Which of the following statements is true?
A.) The momentum before the collision is equal to the momentum after the collision.
B.) The momentum before the collision is greater than the momentum after the collision.
C.) The momentum before the collision is less than the momentum after the collision.
Physics
2 answers:
LuckyWell [14K]2 years ago
8 0
The answer is b.) the momentum before the collision is greater than the momentum after the collision
NikAS [45]2 years ago
8 0

Answer: A.) The momentum before the collision is equal to the momentum after the collision.

Explanation:

From the Law of conservation of momentum, in a closed system, the net momentum is conserved.

Be it elastic collision or inelastic collision, whenever there is no external force acting in a system, the net momentum is always conserved.

Thus, initial momentum before collision is equal to the final momentum after collision.

Correct option is A.

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Wave base is the most affected by a wave in deep water. :]
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3 years ago
Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
What is the equation to find an angle of projectile
Ksju [112]
I do not recall the answer to this question
3 0
2 years ago
The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he
VashaNatasha [74]

Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0
2 years ago
What is the displacement of the object from 2 seconds to 6 seconds?
masya89 [10]

the answer would be 4

8 0
3 years ago
Read 2 more answers
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