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harina [27]
3 years ago
7

11.

Physics
1 answer:
kicyunya [14]3 years ago
6 0

Answer:

Explanation:

From the graph attached,

Values on y-axis represent the distance traveled by a particle at any moment of time 't' given at x-axis.

Velocity of the particle = \frac{\triangle y}{\triangle x}

a). At t = 1 s,

Position of the the point is at (1, 5).

Therefore, slope of the line = velocity of the particle

Velocity = \frac{\text{Distance traveled from x= 0 to x = 1}}{\text{Duration}}

             = \frac{5}{1} m per sec

             = 5 m per sec.

b). At t = 3 s,

Velocity of the particle = Slope of the segment on which the particle lies

Velocity = \frac{10-4}{4-2}

              = 3 m per sec.

c). At t = 4.5 s,

Velocity = 0 [Since distance traveled = 0]

d). At t = 7.5 s

Velocity = \frac{0+6}{8-7}

              = 6 m per sec.

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Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.

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3) connect two lamps to a power supply in series and current drawn from the power supply is is. connect the same two lamps in pa
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The current drawn by the series circuit is(R₁ + R₂)/R₁R₂  times the current drawn by the parallel circuit.

Let the resistance of the two lamps are R₁ and R₂.

Then the equivalent resistance in series combination is: R = R₁ + R₂.

And,  the equivalent resistance in parallel combination is:

r = R₁R₂/(R₁ + R₂).

So, if the supply voltage is V,

Then, current drown in series combination; i_s = V/R = V/(R₁ + R₂)

And, current drown in parallel combination; i_p = V/r = V(R₁ + R₂)/R₁R₂

So ,\frac{i_s}{i_p} = [ V/(R₁ + R₂)] /[V(R₁ + R₂)/R₁R₂]

=  (R₁ + R₂)/R₁R₂

Hence, the  ratio of current drawn in series and current drown in parallel is  (R₁ + R₂)/R₁R₂.  So, he current drawn by the series circuit is(R₁ + R₂)/R₁R₂  times the current drawn by the parallel circuit.

Learn more about electric current here:

brainly.com/question/2264542

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