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vfiekz [6]
2 years ago
14

25. Spider-Man Spider-Man holds the bottom of an elevator with one hand.

Physics
1 answer:
sergij07 [2.7K]2 years ago
4 0

Answer:

Explanation:

a) F = m(g + a) = 50(10 + 0.0) = 500 N

b) F = m(g + a) = 50(10 + 2.0) = 600 N

c) F = m(g + a) = 50(10 - 2.0) = 400 N

d) F = m(g + a) = 50(0.0 + 0.0) = 0.00 N

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
1. the most suitable cutting process to use on ferrous metals such as straight carbon steel is
dangina [55]

Answer:

plasma arc

cutting

Explanation:

It is plasma arc cutting because i think it is

6 0
3 years ago
Read 2 more answers
1. Al descomponerse su vehículo una persona tira de su auto con la ayuda de una cuerda con una fuerza de 3500 N que forma un áng
OleMash [197]

Responder:

Fy = 2474,8737

Fx = 2474,8737

Explicación:

Dado que :

Dado:

Fuerza, F = 3500 N

Ángulo formado con la horizontal, θ, = 45 °

Los componentes de una fuerza se pueden descomponer en componentes verticales y horizontales.

El componente vertical Fy; y

El componente horizontal Fx

Fy = Fuerza * sinθ

Fy = 3500 * sin45 °

Fy = 2474,8737

El componente horizontal:

Fx = Fuerza * cosθ

Fy = 3500 * cos45 °

Fy = 2474,8737

8 0
2 years ago
Wood has chemical energy, which can be used to generate thermal and radiant energy when burned in a fireplace. Which best explai
Marianna [84]

Answer:In the case of burning wood, stored potential energy (in the form of chemical energy) in the log is released due to heating by other excited atoms. This chemical reaction is called combustion and requires oxygen. Combustion changes the potential chemical energy into kinetic energy in the form of heat.

Explanation:

7 0
3 years ago
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Could dark matter be a higher dimensional object? Why?
REY [17]
Well dark matter is , in a layman terms just a concept used to identify missing mass in the universe. (extremely simplified and probably wrong but look that up). It could be but then if it is then we would still be able to percive the matter in some form. All we know from instruments is that they have mass and absorb light. We would need more info before any conclusions but it is 100% a possibility. But remember that when someone says "3-d" for example they are talking about a certain space.. technically matter exists on all planes (at least all the planes we know of in our local universe.) The laws of the cosmos are not yet proven to be constant and even if they are what of outside it?

When you take high school physics ask your teacher.
3 0
3 years ago
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