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BARSIC [14]
3 years ago
10

The position vector of a particle of mass 2 kg is given as a function of time by ~r = (5 m) ˆı + (5 m/s)t ˆ . Determine the mag

nitude of the angular momentum of the particle with respect to the origin at time 7 s
Physics
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

L = 50 \hat{k}\ kg.m^2/s

Explanation:

given,

mass of the particle = 2 Kg

position vector of the particle is given as

r = 5\hat{i} + 5 t \hat{j}

v =\dfrac{dr}{dt}

\dfrac{dr}{dt} = 5 \hat{j}

angular momentum

L = m (r x v)

L = m[(5\hat{i} + 5 t \hat{j}\ )\times (5 \hat{j})]

cross product of vectors

i x j = k

j x j = 0

L = m\times 25 \hat{k}

L = 2\times 25 \hat{k}

L = 50 \hat{k}\ kg.m^2/s

Angular momentum is not dependent on the time function.

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Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

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Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

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Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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D

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kondor19780726 [428]

Answer:

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