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alina1380 [7]
4 years ago
6

Ayo, how do I change my username on here?

Engineering
1 answer:
nydimaria [60]4 years ago
4 0

Answer:

I'm not sure

Explanation:

eeeeeeeeeeeeeeeeeeeeeee

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The primary difference between LEED Certification for Buildings and LEED Professional Certifications is that professionals can c
TEA [102]

Answer:false

Explanation:

8 0
3 years ago
A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th
Dmitrij [34]

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

6 0
3 years ago
What is the lehr and what purpose does it serve?
levacccp [35]

Explanation:

Step1

Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.  

Step2

Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product.  So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.

5 0
3 years ago
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati
Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

\dot m = 5000 kg/s

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy  are taken as zero.

change in mechanical energy is givrn as

e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}

                         = 0.491 kJ/kg

\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW

\eta_{OVERALL}  = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76

B) \eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80

c) \dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)

\dot W_{shaft} = 1964 kW

7 0
4 years ago
One more because i am running out of points! Have a lovely rest of your guyses day! :)
VMariaS [17]

Answer:

thank you

Explanation:

so much for that

8 0
3 years ago
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