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svlad2 [7]
3 years ago
10

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi

dth b 10 mm (0.4 in.); the distance between support points is 45 mm (1.75 in.). Compute the flexural strength if the load at fracture is 290 N (65 lb
Engineering
1 answer:
Anon25 [30]3 years ago
4 0

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

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If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD t
Licemer1 [7]

Answer:

<em>866.1 N</em>

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N   this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = <em>866.1 N</em>

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zubka84 [21]

Answer:

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A homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time. Determine the average cost per k
Llana [10]

Answer:

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Explanation:

Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.

Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.

For the first 100 kWh:

16 cent × 100 = 1600 cents = 16 dollars

Since 1 dollar = 100 cents

For the remaining energy:

260 - 100 = 160 kwh

10 cents × 160 = 1600 cents = 16 dollars

The total cost = 10 + 16 + 16 = 42 dollars

Note that the base monthly of 10 dollars is added.

The cost of 260 kWh of energy consumption in July is 42 dollars

To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.

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Convert it to cents by multiplying the result by 100.

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Explanation:

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