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2 years ago

How may acres of land are estimated to be lost each day to development?

Engineering

1 answer:

Law Incorporation [45]2 years ago

3 0

175 acres per hour of agricultural land lost to development – 3 acres per minute.

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______ are an idication that your vehicle may be developing a cooling system problem.

iris [78.8K]

Answer:

The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is an idication that your vehicle may be developing a cooling system problem.

Explanation:

8 0

2 years ago

The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The

klemol [59]

The power that must be supplied to the motor is 136 hp

Explanation:

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power = ?

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * a$

a = 16.1 ft/s²

We know,

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s$

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

= 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0

3 years ago

The drag force, Fd, imposed by the surrounding air on a

Ad libitum [116K]

Answer:

a) 23.551 hp

b) 516.89 hp

Explanation:

given:

$F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2$

required:

the power in hp

solution:

$(F_{d})_{a} =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}$ .............(1)

by substituting in the equation (1)

=353.27 lbf

$(F_{d})_{b} =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}$ ..........(2)

by substituting in the equation (2)

= 2769.29 lbf

power is defined by

P=F.V

$P_{a}=$ 353.27*36.67

=12954.411 lbf.ft/s

=12954.411*.001818

=23.551 hp

$P_{a}=$ 2769.29*102.67

= 284323 lbf.ft/s

= 284323*.001818

= 516.89 hp

3 0

2 years ago

A reversible compression of 1 mol of an ideal gas in a piston/cylinder device results in a pressure increase from 1 bar to P2 an

Mashutka [201]

Answer:

attached below

Explanation:

6 0

2 years ago

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making Y the subject

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}$

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, $\sigma_c$ is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682$

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making K the subject

$K=\sigma_c Y \sqrt {a\pi}$ and substituting 260 MPa for $\sigma_c$ while a is taken as 0.003m and Y is already known

$K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa$

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0

2 years ago