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2 years ago

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

Answer:

b) 200 N

Hope you could understand.

If you have any query, feel free to ask.

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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

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The length = 20 mm below the centriod

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

$\tau = \dfrac{VQ}{Ib}$

where;

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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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