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marshall27 [118]
3 years ago
11

Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh

xe AB = L ( m) . Ô tô không bị trượt ngang và đang quay vòng trên đoạn đường có góc nghiêng aphal , bán kính quay vòng r ( m ), vận tốc xe v ( m/s ). Tính chiều cao trọng tâm lớn nhất để xe không bị lật ngang .
Engineering
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer:

wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh

Explanation:

jwhwhwhwhwhwwhhahwhahahwh

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Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0
3 years ago
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The "Big Dig" was the nickname of the civil engineering project that redesigned the highway Infrastructure for the city of
zheka24 [161]
Geotechnical since it’s geologicaly based
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3 years ago
A full-adder is a combinational circuit that forms the arithmetic sum of three input bits.
Vinvika [58]
(b) correct it is false
5 0
2 years ago
A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,
juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

-  Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE  as follows:

                                    stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B   and A_ac: cross sectional area of member AE.

                                    cos(Q) = 4 / 5   ..... From figure ( trigonometry )

                                    A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

                                    stress_ae = 2*(4/5) / 3*10^-4

                                    stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

                                   (M)_c = P*6 - F_eb*3

                                      0 = P*6 - F_eb*3

                                      F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

                                      P - F_eb - F_c = 0

                                      F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

                                     shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

                                     A_pin = pi*d^2 / 4

                                     A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

                                     shear stress = 0.5*P / 2*A_pin

                                     shear stress = 0.5*2 / 2*2.8353*10^-4

                                    shear stress = 1.763 MPa

7 0
3 years ago
Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent
Aleksandr [31]

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar \sigma = 540 M pa

Cross section area of the bar = 8.1^{2}  = 65.61 mm^{2}

We know that the ultimate strength of the bar is calculated from

\sigma = \frac{P_{max} }{A}

540 = \frac{P_{max} }{65.61}

P_{max} = 540 × 65.61

P_{max} =  35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

6 0
3 years ago
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