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2 years ago

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A. A cord passing over an easily turned pulley (one that is both massless and frictionless) has 7kg mass hanging from one end an

d a 9kg mass hanging from the other end as shown below. G 7.0 kg 9.0 ks This arrangement is known as Atwood's machine. Calculate the (i). acceleration of the masses (ii). tension in the cord. [2.5

1 answer:

Zepler [3.9K]2 years ago

6 0

Answer:

M g = M a = a (9 - 7) kg with the two masses going in different directions

(7 + 9) kg * g = a * 2 kg 2 kg effective in prducing acceleration

a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2

M1 a = T - M1 * g T producing upward acceleration of 1.23 m/s^2

T = 7 (a + g) = 7 * 11 = 77 N

Check:

M2 a = M2 * g - T M2 producing downward acceleration

T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N

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What effects does using fossil fuels have on our planet

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3 years ago

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. A 65 kg climber on top of Mt. Everest (8800 m high). How much Pontential Energy does he have?

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He has 5,609,404 J of potential energy

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3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

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Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

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Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

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Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

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Explanation:

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