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frez [133]
3 years ago
15

the average velocity of a duck is zero as it goes from t=0 to t=120 s. describe the displacement of the duck for that interval

Physics
1 answer:
podryga [215]3 years ago
8 0

Answer:

The displacement, s= 0 m

Explanation:

Given data,

The average velocity of the duck, v = 0 m/s

The time interval of the velocity, t₁ = 0 and t₂ = 120 s,

                                       ∴                 Δt =  t₂ - t₁

                                                                = 120 s

The displacement of the duck is given by the formula,

                                         s = v x Δt

                                            = 0 m/s x 120 s

                                            = 0 m

Hence, the displacement of the duck is, s = 0 m

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A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
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Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
3 years ago
The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 k
Murrr4er [49]

Answer:

28716.4740661 N

1.2131147541 m/s

51.2474965841%

Explanation:

m = Mass of plane = 74000 kg

s = Displacement = 3.7 m

f = Frictional force = 14000 N

t = Time taken = 6.1 s

u = Initial velocity = 0

v = Final velocity

s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2

Force is given by

F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N

The force with which the team pulls the plane is 28716.4740661 N

v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J

Work done is given by

W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J

The fraction is given by

\dfrac{54450.9540448}{106250.954045}=0.512474965841

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

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