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3 years ago

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A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g 5 9.80 m/s2. Find (a) the period of the pend

ulum and (b) its length.

1 answer:

choli [55]3 years ago

6 0

Answer:

(a) 1.5 second

(b) 0.56 m

Explanation:

Pendulum makes 120 oscillations in 3 min that means in 180 seconds

time taken by the pendulum to complete one oscillation is called time period.

(a) So, the time period is 180 / 120 = 1.5 second

T = 1.5 second

Thus, the time period of the pendulum is 1.5 second.

(b) g = 9.8 m/s^2

The formula for the time period is given by

$T =2\pi \sqrt{\frac{L}{g}}$

Where, L be the length of pendulum

$1.5 =2\times 3.14 \sqrt{\frac{L}{9.8}}$

$0.057= {\frac{L}{9.8}}$

L = 0.56 m

Thus, the length of the pendulum is 0.56 m .

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Answer:

The displacement is 386.16m

Explanation:

A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement

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Since this is a right angle triangle

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R= sqrt(149120)

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Answer:

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Explanation:

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now we know that

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z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

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Answer:Dissociative Identity Disorder

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3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

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Answer:

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lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

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To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

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3 years ago

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3 years ago