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Kamila [148]
3 years ago
10

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

s of 0.425 kg and the player's foot is in contact with it for 5.10 ✕ 10−2 s, find the x- and y-components of the soccer ball's change in momentum and the magnitude of the average force exerted by the player's foot on the ball.
PART A: the x- and y-components of the soccer ball's change in momentum (in kg · m/s)

Δpx ?????????? = kg · m/s

Δpy ?????????? = kg · m/s
Physics
1 answer:
Alexxandr [17]3 years ago
3 0

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

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5. {Chapter 6 Suppose you are traveling down the road in your car at highway speeds on a nice
Hitman42 [59]

Answer:

a) The 2 forces are equal

b) The impulse is the same

c) The change in momentum is the same

d) Inelastic

Explanation:

a)

According to Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify:

- The car as object A

- The bug as object B

Therefore:

- The force exerted by the car on the bug is the action force

- The force exerted by the bug on the car is the reaction force

According to Newton's third law of motion, these two forces are equal in magnitude, and opposite in direction.

b)

The impulse exerted on an object is equal to the product between the force exerted on it and the time taken:

I=F\Delta t

where

I is the impulse

F is the force exerted

\Delta t is the time during which the force is applied

In this problem:

- The force F exerted on both the car and the bug is the same (according to what we said in part a)

- The time interval \Delta t is the same for the two objects

Therefore, the impulse exerted on the bug is the same as the impulse on the car.

c)

The change in momentum of an object is defined as:

\Delta p = m \Delta v (1)

where

m is the mass of the object

\Delta v is the change in velocity of the object

However, according to the impulse theorem, the change in momentum of an object is also equal to the impulse it has experienced:

\Delta p = I

Therefore, since the impulse experienced by the bug and the car is the same (part b), this means that the change in momentum of the bug and the car is the same (and so by looking at eq.(1), we can conclude that the bug will experience a larger change in velocity, since its mass is smaller than that of the car).

d)

There are two types of collision:

  • Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
  • Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of the energy is converted into thermal energy due to the presence of friction forces)

In particular, inelastic collision occurs when the two objects stick together after the collision.

In this situation, we see that the bug and the car stick together after the collision: this means that the total kinetic energy of the system is not conserved, and therefore, this is an example of inelastic collision.

5 0
3 years ago
The electromagnetic wave that delivers a cellular phone callto
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Answer:

E=2.41\cdot 10^{-5} J

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}} (1)

  • c is the speed of light (3*10⁸ m/s)
  • μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
  • B₀ is the magnetic field

I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}

I_{average}=2.68\cdot 10^{-6} W/m^{2}

Now, let's define the relationship between power (P) and average intensity (I).

I_{average}=\frac{P}{A}

  • P is the power
  • A is the area crossed

So we can calculate the power.

P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W

Finally, energy is the product of P times time, so:

E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J

I hope it helps you!

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To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is
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Answer:

Explanation:

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Which statement is true about unltraviolet waves? They are
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Need to see the statements mentioned in the q posted.
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3. What type of current is produced by a battery? (1 point
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Answer:

direct current hope this helped : )

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