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3 years ago

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In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s2 and

9.88 m/s2. Determine (i) the percent difference of the measurements (ii) the percent error of their mean. [Take the theoretical value of g to be 9.81 m/s

1 answer:

ohaa [14]3 years ago

4 0

Answer:

i. +/- 1.43% and +/- 0.71% ii. +/- 0.33%

Explanation:

$% Error = \frac{Error}{Measurement}* 100%\\$

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myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

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

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

$T'=2\pi \sqrt{\frac{L'}{g}}$

$T'=2\pi \sqrt{\frac{9L}{g}}$

Taking out the square root of 9 (3):

$T'=3*2\pi \sqrt{\frac{L}{g}}$

Substituting the original T:

$T'=3*T$

The new period of oscillation is D) 3.0 T

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Mary spins a ball in a horizontal circle on a massless string. Karen has a similar ball but

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3 years ago

A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe

marin [14]

Answer:

a) $T_f=305.7049\ K$

b) $\Delta S=313.51\ J.K^{-1}$

Explanation:

Given:

mass of copper, $m_c=2.64\ kg$

initial temperature of copper, $T_{ic}=400\ K$

specific heat capacity of copper, $c_c=385\ J.kg^{-1}.K^{-1}$

mass of water, $m_w=4\ kg$

initial temperature of water, $T_{iw}=300\ K$

specific heat capacity of water, $c_w=4200\ J.kg^{-1}.K^{-1}$

a)

<u>∵No heat is lost in the environment and the heat is transferred only between the two bodies:</u>

Heat rejected by the copper = heat absorbed by the water

$2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)$

$T_f=305.7049\ K$

b)

<u>Now the amount of heat transfer:</u>

$Q=m_c.c_c.(T_{ic}-T_{f})$

$Q=2.64\times 385\times (400-305.7049)$

$Q=95841.5841\ J$

∴Entropy change

$\Delta S=\frac{dQ}{T}$

$\Delta S=\frac{95841.5841}{305.7049}$

$\Delta S=313.51\ J.K^{-1}$

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4 years ago