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fomenos
3 years ago
8

A quarterback, Patrick, throws a football down the field in a long arching trajectory to wide receiver, Tyreek. The football and

Tyreek are traveling in the same direction, started at the same spot, and the football was thrown at the same instant that Tyreek began running. Furthermore, both Tyreek and the football have horizontal components of speed of 22.6 mph. Under these circumstances, no matter what angle the football is thrown at, it will land on Tyreek (whether he catches it or not). True or false
Physics
1 answer:
irga5000 [103]3 years ago
8 0

Answer:

 the statement is False       \frac{x_{ball} }{ x_{rec} } = cos θ

Explanation:

Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.

Let's start by finding the time it takes for the ball to reach the ground

        y = v_{oy} t - ½ g t²

when it reaches the ground its height is y = 0

       0 = vo sin θ  - ½ g t²

       0 = t (vo sin θ - ½ g t)

The results are

       t = 0                             exit point

       t = 2 v₀ sin θ/g            arrival point

at this point the ball traveled

       x_{ball}= v₀ₓ t

       x_{ball} = v₀ cos θ  2v₀ sin θ / g

       x_{ball}= 2 v₀² cos θ   sin θ/ g

Now let's find that distantica traveled the receiver in time

        x_{rec} = v₀ t

        x_{rec} = v₀ (2 v₀ sin θ / g)

        x_{rec} = 2 v₀² sin θ / g

without dividing this into two distances

          \frac{x_{ball} }{ x_{rec} } = cos θ

therefore the distances are not equal to the ball as long as behind the receiver

Therefore the statement is False

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Answer:

Kinectic friction is the answer to the lower part of the question

5 0
3 years ago
A tennis ball is dropped from 1.16 m above the
sweet-ann [11.9K]

Answer:

Vf = 4.77 m/s

Explanation:

During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 1.16 m

Vf = Final Velocity of Ball = ?

Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)

Therefore, using these values in the equation, we get:

(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²

Vf = √(22.736 m²/s²)

<u>Vf = 4.77 m/s</u>

6 0
3 years ago
An AC generator consists of 20 circular loops of wire with an area of 75 cm2. It has a maximum induced voltage of 24 V. If its a
Monica [59]

Faraday's law allows us to find the magnetic field that produces the emf in the rotating system is:

  • The magnetic field is:  B = 0.424 T

Faraday's law of induction states that when the magnetic flux changes in time, an induced electromotive force is produced.

            fem = - \frac{d \Phi_B }{dt}  

where fem is the induced electromotive force and Ф the flux,

The magnetic flux is the scalar product of the field and the area.

           \Phi_B = B . A = B A  \ cos \theta  

In this case we have several turns, so the expression remains.

           fem = - N B A \ \frac{d cos \theta}{dt}  

Indicate that the turns rotate at a constant frequency, therefore we can use the uniform rotational motion ratio.

           

           θ = w t

We substitute

 

         fem = - N B A \ \frac{d \ cos \ wt}{dt}\\fem =  N B A w sin \ wt

the maximum induced electromotive force occurs when the sine function is ±1

          fem = N B A w

They indicate that the fem = 24 V, the number of the turn is N = 20, the area is A = 75 cm² = 75 10⁻⁴ m² and the frequency f = 60 Hz

Frequency and angular velocity are related.

           w = 2π f

We substitute.

           fem = N B A 2π f

           B = \frac{fem }{2 \pi \ NA \ f}  

Let's calculate.

         B= \frac{24 }{2\pi \ 20 \ 75 \ 10^{-4} 60}B = 24 / 2pi 20 75 10-4 60

         B = 0.424 T

In conclusion, using Faraday's law we can find the magnetic field that produces the emf in the rotating system is:

  • The magnetic field is; B = 0.424 T

Learn more about Faraday's law here:  brainly.com/question/24617581

8 0
3 years ago
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height o
irina [24]

Answer: 20.765 m/s

Explanation:

This problem can be solved by the conservation of energy principle, this means the initial energy E_{o} must be equal to the final energy  E_{f}:

E_{o}=E_{f} (1)

Where each energy is the sum of kinetic energy K and potential energy U:

K_{o}+U_{o}=K_{f}+U_{f} (2)

Where:

K_{o}=\frac{1}{2}mV_{o}^{2}

Being m your mass and V_{o}=0 m/s your initial velocity, since the roller coaster sterted from rest.

U_{o}=mgh_{o}

Being  g=9.8 m/s^{2} the acceleration due gravity and  h_{o}=25 m your initial height

K_{f}=\frac{1}{2}mV_{f}^{2}

Being V_{f} your final velocity

U_{f}=mgh_{f}

Being h_{f}=3 m your final height

Rewritting (2):

\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f} (3)

mgh_{o}=m(\frac{1}{2}V_{f}^{2}+gh_{f}) (4)

Isolating V_{f}:

V_{f}=\sqrt{2g(h_{o}-h_{f})} (5)

V_{f}=\sqrt{2(9.8 m/s^{2})(25 m-3 m)} (6)

Finally:

V_{f}=20.765 m/s This is your spedd when you arrive at 3 m height

7 0
4 years ago
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . how fast wil
Mamont248 [21]
Applying conservation of momentum
Quarterback mass = 80 kg
ball mass = 0.43 kg
Initially both together but horizontal velocity of both 0
initial momentum = 0
Final momentum = 15*0.43 - 80v
initial = final (law of conservation of momentum)
6.45 = 80v
v = 0.08 m/s
6 0
4 years ago
Read 2 more answers
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