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2 years ago

11

a team of engineer's is designing a new Rover to explore the surface of Mars which statement describes the clearest constraint t

hat applies to the solution

2 answers:

nata0808 [166]2 years ago

5 0

Answer:it must operate at temperatures below 0’C

Explanation:

zmey [24]2 years ago

3 0

Answer:

it must operate at temperatures below 0 degrees Celsius

Explanation:

You might be interested in

2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam

Natasha2012 [34]

Answer:

$\eta = 91.7$ %

Explanation:

Determine the initial velocity

$v_1 = \frac{\dot v}{A_1}$

$= \frac{0.1}{\pi}{4} 0.08^2$

= 19.89 m/s

final velocity

$v_2 =\frac{\dot v}{A_2}$

$= \frac{0.1}{\frac{\pi}{4} 0.12^2}$

=8.84 m/s

total mechanical energy is given as

$E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$\dot v = \dot m v$ $( v =v_1 =v_2)$

$E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$= mv\Delta P + \dot m \frac{v_2^2 -v_1^2}{2}$

$= \dot v \Delta P + \dot v \rho \frac{v_2^2 -v_1^2}{2}$

$= 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}$

$E_{mech} = 36.34 W$

Shaft power

$W = \eta_[motar} W_{elec}$

$=0.9\times 44 =39.6$

mechanical efficiency

$\eta{pump} =\frac{ E_{mech}}{W}$

$=\frac{36.34}{39.6} = 0.917 = 91.7$ %

8 0

2 years ago

The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards

lesya [120]

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

$a_r = \frac{V^2}{R}$

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

7 0

2 years ago

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which

Westkost [7]

Answer:

Entropy generation rate of the two reservoirs is approximately zero ( $\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$ ) and system satisfies the Second Law of Thermodynamics.

Explanation:

Reversible heat pumps can be modelled by Inverse Carnot's Cycle, whose key indicator is the cooling Coefficient of Performance, which is the ratio of heat supplied to hot reservoir to input work to keep the system working. That is:

$COP_{H} = \frac{\dot Q_{H}}{\dot W}$

The following simplification can be used in the case of reversible heat pumps:

$COP_{H,rev} = \frac{T_{H}}{T_{H} - T_{L}}$

Where temperature must written at absolute scale, that is, Kelvin scale for SI Units:

$COP_{H, rev} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

$COP_{H, rev} = 17.479$

Then, input power needed for the heat pump is:

$\dot W = \frac{\dot Q}{COP_{H,rev}}$

$\dot W = \frac{300\,kW}{17.749}$

$\dot W = 16.902\,kW$

By the First Law of Thermodynamics, heat pump works at steady state and likewise, the heat released from cold reservoir is now computed:

$-\dot Q_{H} + \dot W + \dot Q_{L} = 0$

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = 300\,kW - 16.902\,kW$

$\dot Q_{L} = 283.098\,kW$

According to the Second Law of Thermodynamics, a reversible heat pump should have an entropy generation rate equal to zero. The Second-Law model for the system is:

$\dot S_{in} - \dot S_{out} - \dot S_{gen} = 0$

$\dot S_{gen} = \dot S_{in} - \dot S_{out}$

$\dot S_{gen} = \frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}}$

$\dot S_{gen} = \frac{283.098\,kW}{280.15\,K} - \frac{300\,kW}{297.15\,K}$

$\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$

Albeit entropy generation rate is positive, it is also really insignificant and therefore means that such heat pump satisfies the Second Law of Thermodynamics. Furthermore, $\dot S_{in} = \dot S_{out}$ .

5 0

2 years ago

Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user

Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

final int NUM_VALS = 4;

int[] userValues = new int[NUM_VALS];

int i;

int matchValue;

int numMatches = -99; // Assign numMatches with 0 before your for loop

matchValue = scnr.nextInt();

for (i = 0; i < userValues.length; ++i) {

userValues[i] = scnr.nextInt();

}

/* Your solution goes here */

numMatches = 0;

for (i = 0; i < userValues.length; ++i) {

if(userValues[i] == matchValue) {

numMatches++;

}

}

System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

}

}

8 0

2 years ago

For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari

Aliun [14]

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = $C$ $(Re_x)^{0.5$ $(Pr)^{1/3$

Nu = $C(\frac{Vx}{v})^{0.5}$ $(Pr)^{1/3$

$\frac{h_xx}{k}$ = $C(\frac{V}{v})^{0.5}$ $(Pr)^{1/3$ $(x)^{0.5$

$h_x$ = $\frac{1}{x^{1/2}}$

Next we write down the expression for the local heat flux from the plate with uniform surface temperature;

q = $h_xA($ T $_s$ - T∞ )

q ∝ $h_x$

∴

q ∝ $\frac{1}{x^{1/2}}$

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

3 0

2 years ago