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2 years ago

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(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

with an initial velocity of 0 m/s. How long does it take for the pumpkin to hit the ground?

1 answer:

slavikrds [6]2 years ago

7 0

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

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Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

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Magnetic field,B=0.9 T

We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

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$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

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When t=0

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$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

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$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

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Answer:

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