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eimsori [14]
3 years ago
6

2 examples of balanced forces

Physics
1 answer:
valina [46]3 years ago
6 0

Answer:

<h2>Here are some examples of situations involving balanced forces. </h2><h2>Hanging objects. The forces on this hanging crate are equal in size but act in opposite directions.</h2><h2>Floating in water. Objects float in water when their weight is balanced by the upthrust from the water.</h2><h2>Standing on the ground.</h2>

I hope this helps

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Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due
prohojiy [21]

Answer:

1.35208 m/s

Explanation:

Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

Time=\frac{100}{0.75}=133.33\ s

Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s

Velocity of the water is 1.125 m/s

From Pythagoras theorem

c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s

So, the man's velocity relative to the shore is 1.35208 m/s

3 0
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