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3 years ago

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You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th

at a pendulum having a length of 35 cm has a period of 2 seconds. What is the acceleration due to gravity on this planet?
A. 13.82 m/s^2
B. 345.4 m/s^2
C. 3.45 m/s^2
D. 1.1 m/s^2

1 answer:

levacccp [35]3 years ago

4 0

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

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

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

$g = \frac{4\pi^2 L}{T^2}$

Our values are given as

$L = 0.35m\\T = 2s\\$

Replacing we have

$g = \frac{4\pi^2 L}{T^2}$

$g = \frac{4\pi^2 0.35}{2^2}$

$g = 3.45 m/s^2$

Therefore the correct answer is C.

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Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

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The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

So

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

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<h2>Yes!</h2>

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At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienc

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Answer:

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Explanation:

Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.

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0 = U^2 - 2 * 40.52 * 9.06

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U = $\sqrt{734.22}$

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Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid

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