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eduard
2 years ago
14

Question 2

Physics
1 answer:
ra1l [238]2 years ago
4 0

Answer:

0.729 meters

Explanation:

d=(V0)t+(1/2)at^2

d=height of shelf

a=9.8 m/s^2 (acceleration due to gravity)

V0=0 m/s (initial downward velocity)

t=0.38586 seconds (1.173 meters/3.04 m/s)

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Which of the following best describes what alveolar are
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Alveoli are tiny balloon shaped structures and are the smallest passageway in the respiratory system. The alveoli are only one cell thick, allowing the relatively easy passage of oxygen and carbon dioxide (CO2) between the alveoli and blood vessels called capillaries.
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If a chemical is completely consumed in a chemical reaction, what is the substance?
Dima020 [189]
Is a process in which One set of substances, called REACTANTS, is converted to a new set of substances is called PRODUCTS.

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5 0
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
Maurinko [17]

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}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

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

(c). We need to calculate the sample's activity

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.

4 0
3 years ago
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kap26 [50]
The frictional force is in the opposite direction
7 0
3 years ago
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

T = 2 \pi \sqrt\frac{m}{k}

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0
3 years ago
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