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Anit [1.1K]
3 years ago
8

People often think that Galileo dropped two objects of dramatically different mass off of the Leaning Tower of Pisa that both hi

t the ground at the same time. Explain why in reality this was most likely not true.
Physics
1 answer:
kramer3 years ago
3 0

Answer: because of air resistance. See explanation for further details.

Explanation: Galileo performed an experiment to proof that the time of descent of two different masses is independent of time.

But in reality this is most likely not true because of air resistance and other fluid frictional effects in consideration.

If the experiment is performed in a vacuum, it will always be true that time is independent of masses of two falling objects.

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Why is it not advisable to touch a canvas tent from inside when it is raining?​
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Answer: When you touch wet canvas, surface tension will draw water to your finger. However, the drop left behind where you touched, like any irregular point on an overhead surface, will draw condensation from inside the tent if it is humid.

3 0
3 years ago
In a fission experiment observed by Hahn and Strassman, uranium-235 was bombarded by neutrons. The products of this ission react
wlad13 [49]

Answer:

helium-4 (90%) or tritium (7%).

Explanation:

hope it helped u buddy

7 0
3 years ago
I really need help.
Zielflug [23.3K]
2/5 = .4
.4*100= 40%
Alex spends more time
4 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
Can someone complete this assignment will give brainliest!<br> The assignment is attached
Musya8 [376]
... I can’t see the attached assignment that you put on here.
3 0
3 years ago
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