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3 years ago

What is the equivalent resistance of the circuit?

Physics

1 answer:

podryga [215]3 years ago

8 0

Answer:

Option D. 5.45 Ω

Explanation:

From the question given above, the following data were obtained:

Resistance 1 (R₁) = 10 Ω

Resistance 2 (R₂) = 20 Ω

Resistance 3 (R₃) = 30 Ω

Voltage (V) = 120 V

Equivalent resistance (R) =?

The equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/10 + 1/20 + 1/30

Find the least common multiple (lcm) of 10, 20 and 30. The result is 60

Divide 60 by each of the denominators and multiply by their numerators respectively. We have:

1/R = (6 + 3 + 2)/60

1/R = 11/60

Invert

R = 60/11

R = 5.45 Ω

Thus, the equivalent resistance in the circuit is 5.45 Ω

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4 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

3 years ago

A bungee jumper has a mass of 60kg and uses a 25m long bungee cord (unstretched length) with an elastic coefficient of 800N/m. a

KonstantinChe [14]

Answer:

a ) 2.68 m / s

b ) 1.47 m

Explanation:

The jumper will go down with acceleration as long as net force on it becomes zero . Net force of (mg - kx ) will act on it where kx is the restoring force acting in upward direction.

At the time of equilibrium

mg - kx = 0

x = mg / k

= (60 x 9.8 ) / 800

= 0.735 m

At this moment , let its velocity be equal to V

Applying conservation of energy

kinetic energy of jumper + elastic energy of cord = loss of potential energy of the jumper

1/2 m V² + 1/2 k x² = mg x

.5 x 60 x V² + .5 x 800 x .735 x .735 = 60 x 9.8 x .735

30 V² + 216.09 = 432.18

V = 2.68 m / s

b ) At lowest point , kinetic energy is zero and loss of potential energy will be equal to stored elastic energy.

1/2 k x² = mgx

x = 2 m g / k

= (2 x 60 x 9.8) / 800

= 1.47 m

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3 years ago

Read 2 more answers