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grandymaker [24]
3 years ago
7

If _____, use threshold braking any time you need to brake hard.

Engineering
2 answers:
Vaselesa [24]3 years ago
8 0

Answer:

You don’t have ABS

Sholpan [36]3 years ago
5 0
Answer: You don’t have ABS
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The simply supported beam in the Figure has a rectangular cross-section 150 mm wide and 240 mm high.
8_murik_8 [283]
Same question idea but different values... I hope I helped you... Don't forget to put a heart mark

4 0
3 years ago
Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
azamat

Answer:

false

Explanation:

3 0
3 years ago
If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
3 years ago
Is an isothermal process necessarily internally reversible? Explain your answer with an example
torisob [31]

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

5 0
2 years ago
Give the approximate temperature (in K) at which creep deformation becomes an important consideration for each of the following
andrezito [222]

Answer:

691K, 543K, 725K, 1473K, 240K, 373K

Explanation:

Creep deformation of any metal is the transformational tendency of a metal to distort rapidly or slowly when attacked by any form of mechanical stress. The temperature significant for a metal to deform is gotten by the division of the actual temperature of the metal by its melting point. This is termed homologous temperature which is 0.4 or higher. It is calculated by the equation:

0.4Tm

Therefore for the listed metals...

For Nickel, 0.4Tm = 0.4 ×(1455 + 273) = 691 K

For Copper, 0.4Tm = 0.4 ×(1085 + 273) = 543 K

For Iron, 0.4Tm = 0.4 ×(1538 + 273) = 725 K

For Tungsten, 0.4Tm = 0.4 ×(3410 + 273) = 1473 K

For Lead, 0.4Tm = 0.4 × (327 + 273) = 240 K

For Aluminium, 0.4Tm = 0.4 ×(660 + 273) = 373 K

5 0
3 years ago
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