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denis23 [38]
3 years ago
9

What is the force on a 1000 kg elevator that is falling freely at 9.8m/s2

Physics
2 answers:
Taya2010 [7]3 years ago
5 0

Explanation:

❀ \underline {{\underline{ \text{Given} }}}:

  • Mass ( m ) = 1000 kg
  • Acceleration ( a ) = 9.8 m/s²

❀ \underline{ \underline{ \text{To \: find}}} :

  • Force ( F )

❀ \underline{ \underline{ \text{Solution}}} :

\boxed{ \sf{force = mass \times acceleration}}

Plug the known values :

⟶\sf{1000 \times 9.8}

⟶ \sf{9800 \: N}

\red{\boxed{ \boxed{ \tt{⟿ \: Our \: final \: answer : 9800 \: N}}}}

Hope I helped !♡

Have a wonderful day / night ! ツ

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

Rainbow [258]3 years ago
4 0

Answer:

Since it is falling freely, the only force on it is its weight, w. w = m ⋅ g = 1000kg ⋅ 9.8m s2 = 9800N To draw a Free Body Diagram, draw an elevator cage (I am sure you would get lots of points for drawing it with intricate detail) with a downward force of 9800 N. I hope this helps,

Explanation:

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Your friend is wearing a red coat. When white light hits the coat, some light is reflected, and some is absorbed.
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Answer:

Orange , yellow, green and blue

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4 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
3 years ago
A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
ElenaW [278]

Explanation:

The given data is as follows.

    Length of beam, (L) = 5.50 m

    Weight of the beam, (W_{b}) = 332 N

     Weight of the Suki, (W_{s}) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

                 \sum M_{o} = 0

     -W_{s} \times x + W_{b} \times 1.5 = 0

                x = \frac{W_{b} \times 1.5}{W_{s}}

                   = \frac{332 N \times 1.5}{505 N}

                   = 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0
4 years ago
1. What are the characteristics of winds?
goldenfox [79]

Answer:

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8 0
4 years ago
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wave
Llana [10]

Explanation:

It is given that,

The thinnest soap film appears black when illuminated with light with a wavelength of 535 nm, \lambda=5.35\times 10^{-7}\ m

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We need to find the thickness of soap film. The soap film appear black means there is an destructive interference. The condition for destructive interference is given by :

2t=m\dfrac{\lambda}{\mu}

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\mu = refractive index

t=m\dfrac{\lambda}{2\mu}

t=\dfrac{\lambda}{2\mu}

For thinnest thickness, m = 1

t=1\times \dfrac{5.35\times 10^{-7}\ m}{2\times 1.33}

t=2.01\times 10^{-7}\ m

Hence, this is the required solution.

6 0
4 years ago
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