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jenyasd209 [6]
3 years ago
15

Acceleration is the rate of change of velocity. The velocity (positive means down) of a pumpkin thrown off the top of Cheadle ha

ll t seconds after launch is 32???? ft/sec (until it hits the ground). (a) What is the average rate of change of velocity between ????=1 and ????=2? 1.5 feet per second per second (b) What is the average rate of change of velocity between ????=1 and ????=1.1? feet per second per second (c) If the pumpkin lands after 2.5 seconds, what is the speed of the pumpkin when it hits? feet per second

Physics
1 answer:
Rom4ik [11]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

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Explain the difference between weight and mass in terms of vectors and scalars
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A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axi
KiRa [710]

Answer:

The induced voltage is  \epsilon  = 4.53 \  V

Explanation:

From the question we are told that

    The  number of turns is  N  =  1300 \  turns

     The diameter is  d  =  2.2 \  cm  =0.022 \ m

     The  initial magnetic field is  B_i  =  0.11 \ T

     The final magnetic field is  B_f  = 0  \ T

    The time taken is  t  = 12 \ ms  =  12*10^{-3} \  s

   

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

         r =  \frac{0.022}{2}

         r = 0.011 \ m

Generally the induced emf  is mathematically represented as

      \epsilon  =  - N *  \frac{d\phi}{dt}

Where  d\phi is the change in magnetic flux of the wire which is mathematically represented as

      d \phi  =  dB*  A *  cos \theta

=>  d \phi  =  (B_f  -  B_i )*  A *  cos \theta

Here  \theta  =  0

since the axis of the coil is parallel to the field

    Where A  is the cross-sectional area of the coil which is mathematically represented as

      A  =  \pi *  r^2

       A  =  3.142 *  0.011^2

      A  =  3.80*10^{-4} \  m^2

So the induced emf

        \epsilon  = -  1300 *  \frac{(0- 0.11) *  3.80*10^{-4}}{12*10^{-3}}   Here we substituted the values of  d \phi

       \epsilon  = 4.53 \  V

6 0
3 years ago
a batsman deflects a ball by an angle of 45 without changing its initial speed which is equal to 54km/h.what is the impulse impa
Andre45 [30]

Answer:

the impulse imparted to the ball is 4.158 kg.m/s

Explanation:

Please find attached image for diagram.

Given;

initial speed of the ball from point A, u = 54km/h = 15 m/s

angle of deflection of the ball along BOP, θ = 45⁰/2 = 22.5⁰

mass of the ball, m = 0.15 kg

the final speed of the ball along B,  = u (in reverse direction)

The initial momentum of the ball, P₁ = mucosθ

The final momentum of the ball is P₂ in reverse direction to P₁.

the impulse imparted to the ball = change in momentum of the ball

J = ΔP = P₂ - P₁

J = mucosθ - (-mucosθ)

J = mucosθ + mucosθ

J = 2mucosθ

J = 2(0.15)15cos(22.5⁰)

J = 4.158 kg.m/s

Thus, the impulse imparted to the ball is 4.158 kg.m/s

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