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Ne4ueva [31]
3 years ago
6

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:
RideAnS [48]3 years ago
8 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.

Stels [109]3 years ago
3 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>

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Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to
sergij07 [2.7K]

Answer:

KE=1.2036\times 10^{-12}\ J

Explanation:

Given:

  • charge on the alpha particle, q=2e=3.2\times 10^{-19}\ C
  • mass of the alpha particle, m=6.64\times 10^{-27}\ kg
  • strength of a uniform magnetic field, B=0.5\ T
  • radius of the final orbit, r=0.5\ m

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

q.v.B=m.\frac{v^2}{r}

m.v=q.B.r

where:

v = velocity of the alpha particle

v=\frac{q.B.r}{m}

v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}

v=1.2048\times 10^{7}\ m.s^{-1}

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m

m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }

m'=6.6533\times10^{-27}\ kg

now kinetic energy:

KE=m'.c-m.c

KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2

KE=1.2036\times 10^{-12}\ J

6 0
3 years ago
A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.
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Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant k=4.45\times 10^3 N/m

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

P_i=\frac{kx^2}{2}

P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J

P_f=mgh=0.25\times 9.8\times h

P_i=P_f

14.24 =0.25\times 9.8\times h

h=\frac{14.24}{0.25\times 9.8}=5.81 m

6 0
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a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v
solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

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The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

                             S = ut + ½ gt²

Let tₓ be the time where the two stones  meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

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The equation for the stone thrown from the base becomes

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Adding these two equations,

                      x + (S - x) = U tₓ

                               S = U tₓ

                               200 = 50 tₓ

∴                                  tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s   

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