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3 years ago

WILL MARK BRAINLIEST!!!!!!!

2 answers:

7 0

In a series circuit, all devices are constrained to a single flow of current. There can only be a single value for the current for all devices, otherwise this would violate the conservation of charge. Therefore the current must be the same across each resistor.

<h3>The answer is D.</h3>

Shkiper50 [21]3 years ago

3 0

Answer:

option D.

Explanation:

In a series circuit, two or more resistors are connected in a line. The equivalent resistance of the circuit is greater than any individual resistor. As more resistors are added, the current decreases. The voltage across each resistor is different. The current remains same across the circuit.

Thus, option D is correct.

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An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

Now the time required for supply of 185370 J:

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago

A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of

Veseljchak [2.6K]

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

$B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}$

Put all the values,

$I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A$

So, it carry current of 2172.46 A.

7 0

3 years ago

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$ (1)

where

F is the magnitude of the force

d is the distance covered by the object

$\theta$ is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

$\theta=0$

And the distance covered is equal to the circumference of the circle, which is:

$d=2\pi r=2\pi (2.5 m)=15.7 m$

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

$W=(3.0)(15.7)(cos 0)=47.1 J$

Learn more about work:

brainly.com/question/6763771

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#LearnwithBrainly

4 0

3 years ago

Pancake syrup is an example of a

laila [671]

It is an example of liquid. if thats what you are asking for...

8 0

3 years ago

When light shines through atomic hydrogen gas, it is seen that the gas absorbs light readily at a wavelength of 91.63 nm. What i

Artist 52 [7]

Answer:

D) 21

Explanation:

When gas absorbs light , electron at lower level jumps to higher level .

and the difference of energy of orbital is equal to energy of radiation absorbed.

Here energy absorbed is equivalent to wavelength of 91.63 nm

In terms of its energy in eV , its energy content is eual to

1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .

Electron is lying in lowest or first level ie n = 1.

Energy of first level

= - 13.6 / 1² = - 13.6 eV.

Energy of n th level = - 13.6 / n². Let in this level electron has been excited

Difference of energy

= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)

13.6 / n² = 13.6 - 13.57 = .03

n² = 13.6 / .03 = 453

n = 21 ( approx )

4 0

3 years ago