Gtjffs

Consider the control volume form of the basic laws. For the conservation of mass form for a control volume with mass flow into a

aleksandrvk [35]

Answer:

C) Dependent on the mass flows in and out.

Explanation:

Lets take control volume(CV)

Take

$m_i$ =inlet mass flow rate

$m_e$ =exit mass flow rate

If we take unsteady flow process then inlet mass can not be equal to exit mass.Some mass can store if inlet mass flow rate is high and exit mass flow rate is low.

So mass of control volume

$m_{cv}=m_i-m_e$ .

so above we can say that mass of control volume dependent on inlet and exit mass.

5 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago

What is the most likely reason the rover won't travel in a straight line?

vovangra [49]

The rovers were designed to trek up to 100 meters (about 110 yards or 328 feet) across the martian surface each martian day, though they have gone much farther. While a complete martian day (called a sol) is about 24 hours and 40 minutes long (or 24 hours 37.5 minutes if you prefer), the Sun can only provide enough power for driving during a four-hour window around high noon. That means the rovers have to be able to move quickly and effectively.
Moving safely from rock to rock or location to location is a major challenge because of the communication time delay between Earth and Mars, which is about 20 minutes on average. Unlike a remote controlled car, the drivers of rovers on Mars cannot instantly see what is happening to a rover at any given moment and they cannot send quick commands to prevent the rover from running into a rock or falling off of a cliff.
During surface operations on Mars, each rover receives a new set of instructions at the beginning of each sol. Sent from the scientists and engineers on Earth, the command sequence tells the rover what targets to go to and what science experiments to perform on Mars. The rover is expected to move over a given distance, precisely position itself with respect to a target, and deploy its instruments to take close-up pictures and analyze the minerals or elements of rocks and soil.

6 0

3 years ago

A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps,

masha68 [24]

Answer:

(a). max possible efficiency = 55.62%

(b). max power output = = 133.5 MW

Explanation:

From the question we were given the Maximum temperature in the system as

Tmax = 500°C

Minimum temperature in the system Tmin = 70°C

the Heat supplied to the boiler Qb = 240000 KJ/s

we use the temperature conversion factor from °C to K

given T(K) = T (°C) + 273

⇒ Tmax = 500 + 273 = 773 K

⇒ Tmin = 70 + 273 = 343 K

(a). we are to determine the maximum possible thermal efficiency;

(Πth)max = 1 - Tmin/Tmax

(Πth)max = 1 - 343/773 = 0.5562

(Πth)max = 55.62%

(b). to determine the maximum possible power output for the plant we have;

(Πth)max = Wmax/Qb

where Wmax rep the maximum power output

(Πth)max = 0.5562

Qb = 240000

∴ Wmax = 0.5562 × 240000 = 133505.6 Kw

Wmax = 133.5 MW

cheers i hope this helps

3 0

3 years ago

Work done by a system during a process can be considered as a property of the system. a)True b) False

SVEN [57.7K]

Answer:

b) False

Explanation:

Work done by a system is not a property because it doesn't define the system's state. Work is mechanical energy exchanged across the system's boundaries.

6 0

4 years ago