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3 years ago

15

In order to avoid a rollover, what is the highest degree incline one should mow on? 10-degree incline 5-degree incline 30-degree

incline 20-degree incline

1 answer:

ser-zykov [4K]3 years ago

6 0

Answer: B: 20-degree incline

Explanation:

A tractor user should avoid slopes of more than 20 degrees in order to avoid rollovers

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The majority of adults now own smartphones or tablets, and most of them say they use them in part to get the news. From 2004 to

noname [10]

Answer:

..... yes

Explanation:

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3 years ago

Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,

77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

$L_{BC}^2 = L_{AB}^2 + L_{AC}^2$

$L_{BC}^2 = (3.5+2.5)^2+ 4^2$

$L_{BC}= \sqrt{(6)^2+ 4^2}$

$L_{BC}= \sqrt{36+ 16}$

$L_{BC}= \sqrt{52}$

$L_{BC}= 7.2111 \ m$

The cross -sectional of the cable is calculated by the formula :

$A = \dfrac{\pi}{4}d^2$

where d = 4mm

$A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2$

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection in length $\delta$ ; we can calculate for the force $F_{BC$ by using the formula:

$\delta = \dfrac{F_{BC}L_{BC}}{AE}$

$F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}$

where ;

E = modulus elasticity

$L_{BC}$ = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for $L_{BC}$ and 0.006 m for $\delta$ ; we have:

$F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}$

$F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN$ ---- (1)

Similarly; we can determine the force $F_{BC}$ using the allowable maximum stress; we have the following relation,

$\sigma = \dfrac{F_{BC}}{A}$

${F_{BC}}= {A}*\sigma$

where;

$\sigma =$ maximum allowable stress

Replacing 190 × 10⁶ Pa for $\sigma$ ; we have :

${F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN$ ------ (2)

Comparing (1) and (2)

The magnitude of the force $F_{BC} = 2.09676 \ kN$ since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

$\sum M_A = 0$

$3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0$

$3.5 P - 3.328 F_{BC} = 0$

$3.5 P = 3.328 F_{BC}$

$3.5 P = 3.328 *2.09676 \ kN$

$P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }$

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

4 0

3 years ago

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

velikii [3]

Answer:

Explanation:

4140-40 I’d pick wood

I hope this helps! :)

4 0

3 years ago

Read 2 more answers

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 . For an applied str

erma4kov [3.2K]

Answer:

B. 2.3mm

Explanation:

The correct answer for the given question is B. 2.3mm. Some aircraft are made are fabricated from aluminium which has plane stain fracture toughness. In the given scenario the plane stain toughness is

Ktc = 40 [ MPa $\sqrt{(m)}$ ]

When the Ktc increases the fracture will appear on the aircraft. In this case the maximum crack until the fracture failure is 2.5mm.

8 0

3 years ago

All MOS devices are subject to damage from:________

Alchen [17]

Answer:

d. all of these

Explanation:

Electrostatic discharge will generally produce excess voltage in a local area that results in excessive current and excessive heat. It will blast a crater in an MOS device, or melt bond wires, or cause damage of other sorts. In short, MOS devices are subject to damage from "all of these."

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3 years ago