PLEASE HELP<br> both questions

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

$Q = ms \Delta T$

$Q = 1*s*(250 - 20)$

$Q = 1*s*230$

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

$Q = ms'\Delta T$

$1*s*230 = 1* s' * \Delta T$

now we have

$\Delta T = (\frac{s}{s'})*230$

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

3 0

3 years ago

Read 2 more answers

Two vehicles A and B accelerate uniformly from rest.

spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

$\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}$

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, $a_A$ = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - $a_A$ ·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, $V_{18A}$ = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, $a_B$ = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ $v_{18B}$ = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, $v_{18B}$ = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle $S_A$ = A₁ + A₂ + A₃

∴ $S_A$ = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle $S_A$ = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, $S_B$ = 440 m

The distance of the two vehicles apart at the 18t second, $S_{AB}$ = $S_B$ - $S_A$

∴ $S_{AB}$ = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, $S_{AB}$ = 60 m.

5 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago

Light incident upon a pane of glass slows down in passing through the glass, Does it emerge at a slower speed or at its initial

Simora [160]

Answer:

It will emerge at its initial speed not a slower speed.

Explanation:

It will emerge at the initial speed because the medium at the point of emergence is the same as the medium before incidence.

Light moves at a constant speed in any particular medium. Hence, the speed of light in air is constant in air and the speed of light in glass is constant in glass.

4 0

3 years ago

BABY BABY BABY OOOHHHH I THOUGHT THAT U WOULD ALWAYS BE MINE

kenny6666 [7]

Answer:

Don't you worry, 'cause everything's gonna be alright, ai-a'ight

Be alright, ai-a'ight

Explanation:

6 0

2 years ago

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PLEASE HELP both questions