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2 years ago

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S A voltage ΔV is applied to a series configuration of n resistors, each of resistance R. The circuit components are reconnected

in a parallel configuration, and voltage ΔV is again applied. Show that the power delivered to the series configuration is 1 /n² times the power delivered to the parallel configuration.

1 answer:

Flura [38]2 years ago

3 0

The power of is series combination is Vn^2 times that of a parallel combination.

For series combination :

Req = R + R + R + ............... n times = nR

I = Δv/nr

Power = (Δv/nr)^2 × nr = Δv^2/nr

For parallel combination

1/req = 1/R + 1/R + 1/R +................(n times) = n/R

Req = R/n

Power = Δv/(R/n) = nΔv^2/R

Ratio = Δv^2/nr/n·Δv^2/R = 1/n^2

Hence, power of is series combination is Vn^2 times that of a parallel.

Learn more about parallel combination here:

brainly.com/question/12400458

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A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is

gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is $\theta = 25 ^o$

$sin \theta = \frac{h}{d}$

Where $h = h_b - h_a$

So $d sin \theta = h_b - h_a$

From the question we are told that

The mass of the block is $m = 20 \ kg$

The mass of the pendulum is $m_p = 2 \ kg$

The velocity of the pendulum at the bottom of swing is $v_p = 15 m/s$

The coefficient of restitution is $e =0.7$

The coefficient of kinetic friction is $\mu _k = 0.5$

The velocity of the block after the impact is mathematically represented as

$v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where $v_2 i$ is the velocity of the block before collision which is 0

$= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15$

Substituting value

$v_2 f = 2.310\ m/s$

According to conservation of energy principle

The energy at point a = energy at point b

So $PE_A + KE _A = PE_B + KE_B + E_F$

Where

$PE_A$ is the potential energy at A which is mathematically represented as

$PE_A = m_b gh_a$ = 0 at the bottom

$KE _A$ is the kinetic energy at A which is mathematically represented as

$K_A = \frac{1}{2} m_b * v_2f^2$

$PE_B$ is the potential energy at B which is mathematically represented as

$PE_B = m_b gh$

From the diagram $h = h_b -h_a$

$PE_B = m_b g(h_b - h_a)$

$KE _B$ is the kinetic energy at B which is 0 (at the top )

Where is $E_F$ is the workdone against velocity which from the diagram is

$\mu_k m_b g cos 25 *d$

So

$\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

Substituting values

$\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$

So

$d = 0.313 \ m$

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3 years ago

An unfortunate 18 kg monkey falls from a 40 m tall tree. What is the monkeys final velocity just befor he impacts the ground.? a

zalisa [80]

The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to $g=9.81 m/s^2$ . The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
$2aS=v_f^2-v_i^2=v_f^2$
from which
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3 years ago

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iogann1982 [59]

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

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3 years ago

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Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

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2 years ago