Which of the following is not a function of the suspension system?____

In RSA Digital Signature, Suppose Bob wants to send a signed message (x = 4) to Alice. The first steps are exactly t eps are exa

Luda [366]

Answer:

what r u on

Explanation:

4 0

3 years ago

How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor

zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0

3 years ago

A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).

Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress $\sigma_m = 90MPa$

stress amplitude $\sigma_a = 190MPa$

a) $\sigma_m =\frac{\sigma_max+\sigma_min}{2}$

$90 =\frac{\sigma_{max}+\sigma_{min}}{2}$ --------------1

$\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}$

$190 = \frac{\sigma_{max}-\sigma_{min}}{2}$ -----------2

solving 1 and 2 equation we get

$\sigma_{max} = 280MPa$

b) $\sigma_{min} = - 100MPa$

c)

stress ratio $=\frac{\sigma_{min}}{\sigma_{max}}$

$=\frac{-100}{280} = -0.35$

d)magnitude of stress range

$=(\sigma_{max} -\sigma_{min})$

= 280 -(-100) = 380 MPa

3 0

3 years ago

Need some help with these plz

pashok25 [27]

100: D, third law of motion

101: D, second law of motion

5 0

3 years ago

If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp

victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0

3 years ago

Which of the following is not a function of the suspension system?_____