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liraira [26]
3 years ago
8

3. The displacement of a particle is represented by s = at? + bts where t is time. Deduce the values of the constants a, and b.

Solution Hint: The summed quantities must all have the same dimension as s. Dimension of displacement s =. L ..., and t = ...I.......... .... ... = Equate dimension of s and at2 ; dim a= . ; dim b = ..... > Equate dimension of s and bt³​
Physics
1 answer:
barxatty [35]3 years ago
5 0

Explanation:

Correct option is

A

As LHS=RHS, formula is dimensionally correct.

Writing the dimensions of either side of the given equation.

LHS=8=displacement=[M

0

LT

0

]

RHS=ut=velocity×time=[M

0

LT

−1

][T]=[M

0

LT

0

]

and

2

1

at

2

=(acceleration)×(time)

2

=[M

0

LT

−2

][T]

2

=[M

0

LT

0

]

As LHS=RHS, formula is dimensionally correct.

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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi
algol13

Answer:

The time taken by the wave to travel  along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire \rho_{s}= 7.86\times10^{3}\ kg/m^{3}

Density of copper wire \rho_{c}=8.92\times10^{3}\ kg/m^3

We need to calculate the linear density of steel wire

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\mu_{c}=\rho_{s}A

\mu_{c}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2

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We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}

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Using formula of velocity

v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}

v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}

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t=t_{s}+t_{c}

t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}

Put the value into the formula

t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}

t=0.458\ sec

t=458\ ms

Hence, The time taken by the wave to travel  along the combination of two wires is 458 ms.

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