You have finished running a marathon and you are relaxing and your body is

vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

$Q_{net,in} - W_{net, out} = \Delta U$ (1)

Where:

$Q_{net, in}$ - Net input heat, measured in joules.

$W_{net, out}$ - Net output work, measured in joules.

$\Delta U$ - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then $\Delta U < 0$ , which is caused by three scenarios:

(i) $Q_{net,in} < 0$ , $W_{net, out} < 0$ , $|Q_{net,in}|>|W_{net,out}|$

(ii) $Q_{net, in} > 0$ , $W_{net,out} > 0$ , $|Q_{net,in}|$

(iii) $Q_{net,in}< 0$ , $W_{net, out}>0$

In the case $Q_{net,in} > 0$ , $W_{net, out}$ , the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

$Q_{net, in} = \Delta U$ , $Q_{net, in} < 0$ (2)

If $Q_{net, in} < 0$ , then $\Delta U < 0$ . Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0

3 years ago

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

3 years ago

What measures air pressure.?

Alekssandra [29.7K]

That depends on what type of pressure you are attempting to measure, to measure Atmospheric pressure, you would use a Barometer. To measure things like tires, you could use a Tire Pressure Gauge. For Industrial processes and boilers, you would use a Manometer. For pressure vessels, you would use a Bordon Gauge.

8 0

3 years ago

Read 2 more answers

How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?

matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be $f_{1}$

So, paper will attain terminal velocity when mg = $f_{1}$

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = $f_{2}$

Rock will attain terminal velocity when Mg = $f_{2}$

Mg > mg

so, $f_{2}$ > $f_{1}$

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

5 0

4 years ago

I didnt want my question public i made a mistake i want it taken down

mrs_skeptik [129]

Then report it and it might be taken down

5 0

3 years ago