1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
olya-2409 [2.1K]
3 years ago
13

An aggregate blend consists of 65% of aggregate A and 35% of aggregate B. The bulk specific gravities of aggregate A and B are 2

.45 and 3.25, respectively. What is the bulk specific gravity of the blend?
a) 2.45
b) 2.68
c) 2.73
d) 2.92
Engineering
1 answer:
lozanna [386]3 years ago
8 0

Answer:

2.68

Explanation:

Percentage by Mass of each Aggregate :

Pa = 65% ; Pb = 35%;

Bulk Specific gravity of each aggregate :

Ga = 2.45 ; Gb = 3.25

Gsb = (Pa + Pb) / (Pa/Ga + Pb/Gb)

Gsb = (65 + 35) / (65/2.45 + 35/3.25)

Gsb = (65 + 35) / 37.299843

Gsb = 100 / 37.299843

Gsb = 2.68

You might be interested in
For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

5 0
3 years ago
Describe the cycle that purifies drinking water. 40 points
Sunny_sXe [5.5K]
Umm the Water cycle sorry I’m trying
5 0
3 years ago
Read 2 more answers
Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate
uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae^{-E/RT} ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

 or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

3 0
3 years ago
Fast plz-The mirror check may involve ______________.
barxatty [35]

Answer:

Realigning the mirror

Explanation:

mirrors should be aligned to minimize blind spots, not look at the tires.

6 0
2 years ago
A large retirement village has a total retail employment of 120. All 1600 of the households in this village consist of two nonwo
polet [3.4K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
Other questions:
  • Nc3
    12·1 answer
  • A 12-mm-diameter cylindrical jet of water discharges vertically into the ambient air. Due to surface tension the pressure inside
    14·1 answer
  • A driver is traveling at 52 mi/h on a wet road. an object is spotted on the road 415 ft ahead and the driver is able to come to
    5·2 answers
  • Module 42 Review and Assessment
    7·1 answer
  • Ayuda con este problema de empuje y principio de arquimedes.
    6·1 answer
  • The sports car has a weight of 4900 lblb and center of gravity at GG. If it starts from rest it causes the rear wheels to slip a
    13·1 answer
  • What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent leng
    9·1 answer
  • Phân tích phương pháp gia công plasma
    10·1 answer
  • Introduction for site visit​
    13·1 answer
  • Describe two fundamental reasons why flexural strength should depend on porosity
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!