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Digiron [165]
2 years ago
10

In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

1°C. How much energy was transferred to the water? (Use a value of 9.8 m/s2 for g.)
Engineering
1 answer:
Orlov [11]2 years ago
6 0

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

6 0
3 years ago
The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc
madreJ [45]

Answer:

Option B

116 ft/s^{2}

Explanation:

\theta=2 rev=2(2\pi)=4\pi

\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})

\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})

\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-4^{2})

\omega_f=8.14 rads/s

v=r\omega=1.75*8.14=14.245 ft/s

Centripetal acceleration =\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}

Tangential component=dr=2*1.75=3.5

Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}

5 0
3 years ago
) A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging proc
bonufazy [111]

Answer:

it will be a scattered plot

Explanation:

5 0
2 years ago
How are project deliverables determined?
Greeley [361]

Answer:

The essence including its problem is listed throughout the clarification section following.

Explanation:

Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.

For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:

  • It should be within the development of the work.
  • The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.

So that the above seems to be the right answer.

7 0
3 years ago
Read 2 more answers
2. What is the original length of the rectangular bar if the deformation is 0.005 in with a force of 1000 lbs and an area of 0.7
Ugo [173]

Answer:

18.75in

Explanation:

Modulus of elasticity = Stress/Strain

Since stress = Force/Area

Given

Force = 1000lb

Area = 0.75sqin

Stress = 1000/0.75

Stress = 1333.33lbsqin

Strain

Strain = Stress/Modulus of elasticity

Strain = 1333.33/5,000,000

Strain = 0.0002667

Also

Strain = extension/original length

extension = 0.005in

Original length = extension/strain

Original length = 0.005/0.0002667

Original length = 18.75in

Hence the original length of the rectangular bar is 18.75in

6 0
2 years ago
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