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marysya [2.9K]
1 year ago
13

prove displacement of a body in nth second is given by the S=u+a/2(2n-1) where symbols have their usual meaning

Physics
1 answer:
Maru [420]1 year ago
7 0

The proof that the displacement of a body in nth second which is given by the S=u+a/2(2n-1) is explained below.

<h3>Calculations and Parameters</h3>

Given:

  • u= initial velocity
  • a= acceleration

Displacement during nth second= displacement in n seconds- displacement in (n-1) seconds

= [un + \frac{1}{2} gn^2] - [ u(n-1) + \frac{1}{2} g(n-1)^2]

= (u * 1) + g(n *1) - \frac{g}{2}(1 *1)

Hence, in (n-1) seconds, n has a unit of time and 1 has a unit of time (not a constant)

Therefore, u(m/s) is multiplied by 1  (unit of time)

g(m/s^2) is multiplied by ( n*1 ), having units of time, i.e s^2

g/2 is also multiplied by ( 1 * 1, also having units of time,  i.e s^2

Therefore, the equation is dimensionally correct and we should note that 1 is not a constant in the equation as the time of 1 second in (n-1) second.

Read more about displacement here:

brainly.com/question/2109763

#SPJ1

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An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
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