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BartSMP [9]
3 years ago
7

Principle of electroplating​

Physics
2 answers:
frosja888 [35]3 years ago
4 0

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uysha [10]3 years ago
3 0

Answer:

Electroplating is the method of depositing one metal over another in the presence of a metal salt (in aqueous solution). The water molecule is released as the final product in this process. As a consequence, electroplating is based on the theory of hydrolysis.

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What is the weight of an object when the object has a mass of 22kg
Savatey [412]
Assuming the object is on earth the objects weight would be equal to its mass multiplied by the gravitational field constant

mass=22kg
g=9.80665N/kg

weight=(22 kg) (9.80665 N/kg)=215.7463N

generally g is rounded to be 10 N/kg so for any question where it asks the weight given the mass just multiply by 10 and that should suffice. In this case the answer would be 220 N
5 0
3 years ago
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie
Nutka1998 [239]

Answer:

i = 0.5 A

Explanation:

As we know that magnetic flux is given as

\phi = NBA

here we know that

N = number of turns

B = magnetic field

A = area of the loop

now we know that rate of change in magnetic flux will induce EMF in the coil

so we have

EMF = NA\frac{dB}{dt}

now plug in all values to find induced EMF

EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})

EMF = 0.2 volts

now by ohm's law we have

current = \frac{EMF}{Resistance}

i = \frac{0.2}{0.40} = 0.5 A

5 0
3 years ago
Which type of circuit is shown?
ivann1987 [24]

Answer:

I think it's D. open series circuit .

Explanation:

<em>hope</em><em> it</em><em> helps</em><em> you</em><em>!</em>

7 0
3 years ago
when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of
Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q​</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

or

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂                                 (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂                            

or, C₂=2X/m. T₂                                   (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂  /2X  /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0
4 years ago
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
3 years ago
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