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lyudmila [28]
3 years ago
9

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​
Physics
1 answer:
Ahat [919]3 years ago
8 0

<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

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Answer:

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Explanation:

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3 years ago
A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0.2 kg, and the robo
Natali5045456 [20]

Answer:

<h3>1.43m/s²</h3>

Explanation:

According to newtons second law.

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5 0
3 years ago
to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to
givi [52]

Answer:

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

f_c = 10 Hz represent the corner frequency

f= 60 Hz represent the original frequency

n represent the filter order and that's the variable that we need to find

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0
3 years ago
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