All categories

1 year ago

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

Physics

1 answer:

Likurg_2 [28]1 year ago

8 0

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

$Q_H$ , $Q_C>0$

You might be interested in

Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam

zhannawk [14.2K]

The velocity of Ned as measured by Pam is the interpretation of v.

Answer: Option D

Explanation:

According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:

The Ned reference framework : (x, t)

The Pam reference framework : $\left(x^{\prime}, t^{\prime}\right)$

From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.

At that point, $v^{\prime}$ is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.

3 0

1 year ago

A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.

Sholpan [36]

Answer:

5cm east& 1cm west from A

Explanation:

https://brainly.ph/question/2753392

7 0

1 year ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

1 year ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

Answer:

Cannonball will be in flight before it hits the ground for 2.02 seconds

Explanation:

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

1 year ago

How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue t

hodyreva [135]

Answer:

The heat transferred from water to skin is 6913.5 J.

Explanation:

Given that,

Weight of water = 25.0 g

Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.

We need to calculate the heat transferred from water to skin

Using formula for stream

$Q=mc\Delta T$

Put the value into the formula

$Q=25\times10^{-3}\times4190\times(373-307)$

$Q=6913.5\ J$

Hence, The heat transferred from water to skin is 6913.5 J.

3 0

2 years ago