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sergey [27]
2 years ago
14

The velocity of 200kg object changes from 10m/s to 22m/s in 12 sec.what is the force on the accelerated object?

Physics
1 answer:
Vikentia [17]2 years ago
4 0
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 22 m/s
  • Time (t) = 12 s
  • Mass (m) = 200 Kg
  • Let the acceleration be a.
  • By using the equation of motion,

v = u + at, we have

  • 22 m/s = 10 m/s + 12 s × a
  • or, 22m/s - 10 m/s = 12 s × a
  • or, 12 m/s = 12 s × a
  • or, a = 1 m/s^2
  • Let the force be F.
  • We know, F = ma
  • Therefore, the force on the accelerated object (F)
  • = ma
  • = (200 × 1) N
  • = 200 N

<u>Answer</u><u>:</u>

<u>b)</u><u> </u><u>2</u><u>0</u><u>0</u><u> </u><u>N</u>

Hope you could understand.

If you have any query, feel free to ask.

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The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th
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Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L  = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

Explanation:

Given that :

The internal shear force V  =  80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress  \tau can be calculated by using the equation:

\tau = \dfrac{VQ}{Ib}

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10  mm

From the centroid ;

Q = Q_1 + Q_{2}

Q = A_1y_1 + A_{2}y_{2}  

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5)  ) mm³

Q = ( 38500 +  307125 ) mm³

Q = 345625 mm³

\tau_H = \dfrac{VQ}{Ib}

\tau_H = \dfrac{80*10^3  * 345625}{64900000*10 }

\tau_H = \dfrac{2.765*10^{10}}{649000000 }

\tau_H = 42.60400616 \ N/mm^2

\mathbf{\tau_H \approx  42.604 \ N/mm^2}

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

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Answer:

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Explanation:

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The density of backpack, d = 30 g/mL

The volume of the backpack, V = 12.3 cm³

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