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Serjik [45]
4 years ago
6

An example of Ferrous alloy is Brass a)-True b)-False

Engineering
1 answer:
djyliett [7]4 years ago
5 0

Answer: False

Explanation: No, brass is not a ferrous alloy.  

      Ferrous alloys are those alloy which contain iron like cast iron, steel, strain-less steel, high carbon steel. Brass on the other hand does not contain any composition. of iron hence it can not be considered as a ferrous alloy. Brass comes under the category of non- ferrous made with a composition of copper and zinc, however their proportion is not strict and we can add other elements like aluminium or lead to alter its durability or corrosiveness.  

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if you help then I will thank u by sooo much I will give tons of points but the answer has to be right.
romanna [79]

Answer:

The first one A, " These are lines of equal air pressure".

Explanation:

8 0
3 years ago
Read 2 more answers
Oliver is designing a new children’s slide to increase the speed at which a child can descend. His first design involved steel b
AVprozaik [17]

Answer:

The correct option is;

A) Steel becomes too hot in the Sun and can burn the children

Explanation:

The properties of steel includes;

Low specific heat capacity, high thermal and electrical toughness, high hardness, high tensile strength, high yield strength, appreciable elongation, high fatigue strength, can easily corrode, high malleability and ability to creep

Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact

8 0
3 years ago
Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)
meriva

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

E(t)=E_o[1-a\frac{T}{T_m}]

where,

E(t) is the modulus of elasticity at any temperature 'T'

E_o is the modulus of elasticity at absolute zero.

T_{m} is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

6 0
3 years ago
A refrigerator operates on average for 10.0 hours an day. If the power rating is the refrigerator is 709 w how much electrical e
Varvara68 [4.7K]

Answer:

61257.6 kW per day

Explanation:

The rating of the refrigerator is given as : 709 w

This means the power consumption is 709 joules per second

The operating time is given as 10 hours.

Change hours to second as

1 hour = 3600 seconds

10 hours = 36000 seconds

Apply the rating as;

709 w = 1 s

?         = 36000 seconds

perform cross product to get;

709 * 36000= 25524000 w

25524000 = 25524kW

For a day, 24 hours will be;

25524 kW = 10 hours

?                 =24 hours

={24*25524}/10 = 61257.6 kW per day

6 0
3 years ago
A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft
tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}

f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}

f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}

          = 0.877

v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}

       = 1,555 veh/hr/lane

FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}

      = (55 + 5) - 6.6 - 0.4 -3 -0

      = 50 mi/h

D = \dfrac{V_P}{s}

D = \dfrac{1555}{55} =28.27

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0
4 years ago
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