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nikklg [1K]
3 years ago
6

Some liquid is poured into a burrete so that it reads 14cm³.50 drops were run each of volume 0.1cm³ .

Physics
1 answer:
nika2105 [10]3 years ago
5 0

Given :

Liquid is poured into a burrete so that it reads 14cm³.

50 drops were run each of volume 0.1cm³ .

To Find :

The volume of liquid in burrete after 50 drops.

Solution :

Volume of each drop, v = 0.1 cm³.

Initial volume in burrete, V = 14 cm³.

Now, volume left after droping 50 drops are :

L = V - 50v\\\\L = 14 - 50\times 0.1 \ cm^3 \\\\L = 9 \ cm^3

Therefore, the volume left in burrete is 9 cm³ .

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Answer:

Approximately 8.4 \times 10^{2}\; \rm N, assuming that g = 9.8\; \rm m \cdot s^{-2}.

Explanation:

Let m and a denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

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  • Upward tension force from the strand of web F(\text{tension}).

The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of F(\text{tension}) (which points upwards) should be greater than that of W (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

(\text{Net Force}) = F(\text{tension}) - W.

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

(\text{Net Force}) = m \cdot a.

Combine these two equations to get:

m \cdot a = (\text{Net Force}) = F(\text{tension}) - W.

Therefore:

\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}.

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately 8.4\times 10^{2}\; \rm N (downwards.)

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