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Yakvenalex [24]
4 years ago
9

A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationsh

ip between the direction of the velocity and acceleration vectors? Assume air resistance is negligible.
A)A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationship between the direction of the velocity and acceleration vectors? Assume air resistance is negligible.
B) At the maximum height, the velocity and acceleration vectors are parallel to each other.
C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other.
D)At the maximum height, the velocity and acceleration vectors point in opposite directions.
E)At the maximum height, both the velocity and acceleration vectors are zero
Physics
2 answers:
zhuklara [117]4 years ago
7 0

At position of maximum height we know that the vertical component of its velocity will become zero

so the object will have only horizontal component of velocity

so at that instant the motion of object is along x direction

while if we check the acceleration of object then it is due to gravity

so the acceleration of object is vertically downwards

so it is along y axis

so here these two physical quantities are perpendicular to each other

so correct answer would be

<em>C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other. </em>

leva [86]4 years ago
6 0

Answer: C) At the maximum height, the velocity and acceleration vectors are perpendicular to each other.

Explanation: At the maximum height the ball stops going up and starts going down. Then, there is a given time at with the velocity of the ball (in the y-axis) is equal to zero.

And as you may know, every lifted object is pulled down by the gravitational force, thie means that the ball is accelerated when it is in the air, so the acceleration is not zero at the maximum height.

Here remember that the player kicked the ball in a given direction, so the ball has a compoent of velocity in the y-axis (that is zero at the point of maximum height) and a component in the x-axis.

Then, at maximum height the velocity is only in the x-direction, and the acceleration is pointing down (in the negative y-direction) so at this point the acceleration and the velocity are perpendiculars.

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aivan3 [116]

Answer:

30.95°

Explanation:

We need to define the moment of inertia of cylinder but in terms of mass, that equation say,

A=\frac{1}{12}m(3r^2+l^2)

Replacing the values we have,

A=\frac{1}{12}(500)(3(0.5)^2+(2)^2)A=197.9kg.m^2

At the same time we can calculate the mass moment of intertia of cylinder but in an axial way, that is,

c=\frac{1}{2}mr^2

c=\frac{1}{2}(500)(0.5)^2

c=62.5kg.m^2

Finally we need to find the required angle between the fixed line a-a (I attached an image )

\Phi = 2tan^{-1}\sqrt{\frac{(\frac{A}{cos\gamma})^2-A^2}{c^2}}

Replacing the values that we have,

\Phi = 2tan^{-1}\sqrt{\frac{(\frac{197.9}{cos5\°})^2-197.9^2}{62.5^2}}

\Phi = 2tan^{-1}(\sqrt{0.076634})

\Phi = 2tan^{-1}(0.2768)

\Phi = 2(15.47)

\Phi = 30.95\°

4 0
4 years ago
Materials have unique properties because each one is made up of different kinds of which particle?
4vir4ik [10]

Answer:

atoms

Explanation:

All  matter is made of elementary particles called "atoms".

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Question 5 of 5: Someone texting or talking spans an average of 27 seconds after they put the phone down are still thinking abou
AVprozaik [17]
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7 0
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India used 600 joules of work on a steel block and moved it 4.8 meters. How much forced was used ?
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8 0
3 years ago
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

  • the peripheral velocity that is directed downward (-V_y) along the y-axis
  • the linear velocity (V_x) that is directed along the x-axis

Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

Also,

-V_y = R* \omega

where \omega(angular velocity) = \frac{d\theta}{dt} = \frac{d}{dt}(8t^4)

-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

3 0
4 years ago
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