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2 years ago

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

1 answer:

3 0

$d = u \times t \: + \frac{1}{2} \times a \times {t}^{2}$
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2

$d = 0.5 \times 9.8 \times {4.2}^{2}$
on solving we get

d = 86.436 metres

Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2

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What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

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