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Brut [27]
4 years ago
11

A football game begins with a coin toss to determine who kicks off. The referee tosses the coin upward with an initial speed of

5 m/s, ignore air resistance.
a)How high does the coin go above the point of release?
b)What is the total time the coin is in the air before returning to the release point?
Physics
1 answer:
erastova [34]4 years ago
5 0

Answer:

(a) 1.28 m

(b) 0.51 s

Explanation:

Given:

Initial speed, U = 5 m/s

Consideration;

1. When the coin is tossed upward, the coin is moving against gravity. Therefor, the acceleration due to gravity experienced by the coin is negative, -g = -9.8 m/s².

2. The final velocity of the coin in the air is zero.  

From equation of motion under free fall,

                           V² = U² + 2gh

                            h = h = \frac{V^{2} - U^{2}}{2(-g)}

Where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

h is the height

                              h = \frac{0^{2} - 5^{2}}{2(-9.8)}

                              h = \frac{0^{-25}{19.60}

                              h = 1.2755 m

                              h = 1.28 m  

(b) The coin still has its initial velocity, u to be equal to 5 m/s, final velocity, v to be equal to 0 m/s and acceleration due to gravity to be -9.8 m/s while in the air before returning to the release point.

From equation of motion under free fall,

                                     V = U + gt

                                     t = \frac{V - U}{g}

                                     t = \frac{0 - 5}{-9.8}

                                     t = 0.51 s

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Answer:

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Explanation:

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now we know that

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d = \frac{v_o + v_1}{2}(80)

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now we can say

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The question is incomplete. Here is the complete question.

A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.

Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.

Answer: Net Force = 50.215.10^{-7}N

Explanation: Force and Magnetic field are related through the following formula:

F = I.L.B.sinθ

Magnetic field (B) in a straight long wire is given by

B=\frac{\mu_{0}.I}{2.\pi.r}

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I is current in the wire;

r is distance to the wire;

Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.

So, for the net force, the relevant forces will be on the sides parallel to the wire.

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F = I.L.B

Replacing magnetic field:

F = \frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}

Note: The side closest to the wire is F₁, while the farthest is F₃.

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F₁ = 278.975*10^{-7}N

Current in F₃ is flowing thoruhg the negative side of the referential, so:

F₃ = -\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}

F₃ = -228.76*10^{-7}N

<u>Net</u> <u>force</u> is total force:

F_{net} = F_{1}+F_{3}

F_{net}=(278.975-228.76).10^{-7}

F_{net}=50.22.10^{-7}

The total force acting on the square loop is F_{net}=50.22.10^{-7}N.

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