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2 years ago

2. In a series circuit, all resistors have identical currents.

Physics

1 answer:

Darina [25.2K]2 years ago

4 0

Answer:

Explanation:

In series connection of resistors, same current flows in the circuit.

Power dissipated by a resistor is

P = i²R

And since same current flows in them, it implies that the power is directly proportional to Resistance, so the higher the resistance of the resistor the higher the power dissipated in the series connection. Also, the lower the resistance, the lower the power dissipated.

2. In parallel connection, same voltage is applied across the resistor.

So, power dissipated by each resistor is

P = V² / R

So, since the same voltage is applied across parallel connection, then, power dissipated in each resistor is inversely proportional to the resistance.,

So, the higher the resistance, the lower the power and the lower the resistance, the higher the power.

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

2 years ago

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

2 years ago

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ololo11 [35]

1.add the amount of the diagram which is M+Y then dived the answer you get.

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2 years ago

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Suppose a ray of light traveling in a material with an index of refraction n a reaches an interface with a material having an in

KATRIN_1 [288]

Answer: C and D

Explanation: One of the first rule for total internal reflection to occur is that the ray must move from a dense to a less dense medium, hence refractive index of medium a must be greater than that of b.

When a ray moves from a dense to a less dense medium, the refracted ray moves away from the normal thus increasing the size of the angle of refraction (total internal refraction occurs when the angle of refraction is 90° and the angle of incidence at this point is known as the critical angle), hence the angle of incidence must be greater than the critical angle.

These points verifies option C and D

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2 years ago

FOR 50 POINTS! -- Your car is parked at the high school. After school you take off and travel for 27 km towards Iowa City. At th

Dovator [93]

Answer:

36km

Explanation:

Im pretty sure displacment is the start and finish in a straight line

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2 years ago

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