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3 years ago

How do you think the temperature difference between the beakers relates to the rate of heat transfer?

Physics

1 answer:

Bogdan [553]3 years ago

8 0

Answer and Explanation:

The rate (in W) at which heat energy moves from the more blazing article to the colder item increments with the temperature distinction between the items.

The more noteworthy the temperature contrast, the more prominent the rate at which exchange of heat energy takes place.

Thus the temperature difference between the beakers relates in the same manner as described above.

The exchange of heat energy continues until a thermal equilibrium is attained.

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How much is a city of a body when it covers 600m in 5 min?

Fofino [41]

Answer:

2m/s

Explanation:

5min × 60sec

=300

now,

600÷300

3 0

3 years ago

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

$T\ cos\theta-mg=0$

$T=\dfrac{mg}{cos\theta}$

Sum of forces in x direction,

$T\ sin\theta=\dfrac{mv^2}{r}$

$mg\ tan\theta=\dfrac{mv^2}{r}$ .............(1)

Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

$T=\dfrac{2\pi r}{v}$

Put the value of T from equation (2) to the above expression:

$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Hence, this is the required solution.

4 0

3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

3 years ago

A force of 16 N to the west is applied to each object below. Which object will

Montano1993 [528]

41kg object that is moving east at 5 m s

7 0

3 years ago

Read 2 more answers

When two or more capacitors are connected in series across a potential difference:

34kurt

Answer:

A) and B) are correct.

Explanation:

Let's take a look at the attached picture. Now

The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE

The equivalent capacitance is going to be: $\frac{1}{C_{total}}=\frac{1}{C_1} +\frac{1}{C_2}$

And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE

And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE

7 0

3 years ago