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sammy [17]
3 years ago
10

Which of the following are equivalent units?

Physics
2 answers:
bezimeni [28]3 years ago
4 0

Answer: B) a newton-meter and a joule

Explanation:

Joule is the unit if energy. One joule is equal to one Newton-meter.

1 J = 1 N-m

Energy is equivalent to work done which is equal to dot product of force and displacement.

E = W = F.s

Energy is measured in Joules. Force is measured in Newtons and displacement is measured in meters.

1 Joule = 1 kg m²/s²

1 N = 1 kg m/s²

1 N-m = 1 kg m/s² × m = 1 kg m²/s² = 1 J

asambeis [7]3 years ago
3 0

Answer : The correct options are, (B) a newton-meter and a joule and (C)  a newton and a joule/sec

Explanation :

As we know that,

1N=1Kg\text{ m }s^{-2}\\\\1J=1\text{Kg }m^2s^{-2}\\\\1W=1\text{Kg }m^2s^{-3}

or we can say that,

1W=1J/s\\\\1J=1Nm\\\\1N=1J/s\\\\1W=1Nms^{-1}

From the given options, the option B and C are the equivalent units.

Hence, the correct options are, (B) and (C)

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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
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Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

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