Physics, 27.01.2021 14:00 kathlynskare06

Please help Will mark brainliest. Spiderman, on a strand of web, accelerates upward at 1.3 m/s2. If spiderman has a mass of 76 kg,
what is the force in the strand of webbing?

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Answer from: anazia10

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

Explanation:

Let and 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 (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.)

Answer from: Quest

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

Answer from: Quest

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

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