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2 years ago

Please help!!!! Will mark brainliest.

Physics

1 answer:

julia-pushkina [17]2 years ago

4 0

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:

Downward gravitational attraction from the earth: $W = m \cdot g$ .
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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Answer:

2. at the lowest point

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2. at the lowest point

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3 years ago

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2 years ago

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In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin

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1.3 x 10⁻⁴ m

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Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

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Answer:

g(h) = g ( 1 - 2(h/R) )

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Hi!

We can derive this expression for distances h small compared to the earth's radius R.

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$F = G \frac{m_1m_2}{r^2}$

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mars1129 [50]

Hola!

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