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2 years ago

One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

Physics

1 answer:

Anettt [7]2 years ago

7 0

The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

N - $W_m -w_{table}$ = 0

N = $W_m +w_{table}$

N = $M_m g + w_{table}$

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

N = 65 9.8 + 100

N = 737 N

c) To calculate the mass of the table we use the relation

W = $m_{table}$ g

$m_{table} = \frac{w_{table}}{g}$

$m_{tabble}= \frac{100}{9.8}$

$m_{table}$ e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Learn more here: brainly.com/question/19860811

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Answer:

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Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

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The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

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$\dfrac{dv}{dt} = 12t+10$

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$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

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$s_{0}=0+0+0+C$

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Additional explanation:

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