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
pickupchik [31]
4 years ago
5

I'm not sure if its true

Engineering
2 answers:
ANTONII [103]4 years ago
4 0

Answer:

the answer is true. your welcome.

shtirl [24]4 years ago
4 0

Answer:

it is true i believe

really all you need to do for this is look at a map

You might be interested in
What is the maximum volume flow rate, in m^3/hr, of water at 15.6°C a 10-cm diameter pipe can carry such that the flow will be l
WITCHER [35]

Answer:

Q = 0.00017597 m3/sec

Explanation:

given data:

temperature of water 15.6 degree celcius

diameter = 10cm =0.1 m

dynamix viscosity of water = .0011193 kg/m sec at 15.6 degree celcius

density of water at 15.6 degree = 999.07 kg/m3

for laminar flow REYNOLD NUMBER < 2000

maximum flow rate =area*velocity

Reynold number Re = \frac{\rho vd}{\mu}

v = \frac{Re \mu}{\rho d}v = \frac{2000*0.00011193}{999.07 *0.1}

v = 0.022406 m/sec

Q =\frac{\pi}{4} d^2 v

where Q is flow expressed in m^3/s

Q =\frac{\pi}{4} *0.1^2 *0.022406

Q =0.00017597 m3/sec

7 0
3 years ago
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
Why is it important to review your plan when you write an algorithm?
user100 [1]
An algorithm is itself a general step-by-step solution of your problem. ... The most important point here is that you must use algorithms to solve problem, one way or the other. Most of the time it's better to think about your problem before you jump to coding - this phase is often called design.
7 0
3 years ago
Question 3. Assign boston_under_10 and manila_under_10 to the percentage of rides that are less than 10 minutes in their respect
Sever21 [200]

Answer:

Enter the following code  to get the required conditions for the answer.

boston_under_5_height = 1.2

manila_under_5_height = 0.6

boston_5_to_under_10_height = 3.2

manila_5_to_under_10_height = 1.4

boston_under_10 = boston_under_10 = 5*boston_under_5_height + 5*boston_5_to _under_10_height

manila_under_10 = manila_under_10 = 5*manila_under_5_height + 5*manila_5_to _under_10_height

Explanation:

Kindly note that question in complete as it belong to the topic of table manipulation and visualization topic. The question asks about the company called Uber and their data extraction from the website called movements.uber.com where data was extracted for 200,000 weekdays in the respective cities of Manila, Philippines and Boston, Massachusetts. Images attached contains the histograms generated for the rides in manila and boston.

3 0
4 years ago
A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Det
suter [353]

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, W_d_{(field)} = 18 kN/m³

maximum dry unit weight measured, W_d_{(max)} = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

RC = \frac{W_d_{(field)}}{W_d_{(max)}}

substitute the given values;

RC = \frac{18}{17} = 1.0588

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

6 0
4 years ago
Other questions:
  • 2. Compare an object
    9·1 answer
  • A stainless steel cap covers a drain in the dolphin tank which is 15 ft deep. The diameter of the hole is 3 in and the cap and w
    12·1 answer
  • A classroom that normally contains 40 people is to be air-conditioned with window air-conditioning units of 5 kW cooling capacit
    6·1 answer
  • Plz help me
    12·1 answer
  • 1. (1 points) What is the name of the drinking water supply well? a. VA1; b. VA24; c. VA19; d. VA40; e. VA18; 2. (1 points) What
    11·1 answer
  • Comparación de hipotecas Los Chos
    15·1 answer
  • How is tolerance calculated?
    11·1 answer
  • How to measure the quality of the output signal in ADC?
    13·1 answer
  • Is Tesla French, American, German, or Russian?
    10·2 answers
  • 1. a major reason for the projected shortage of trained craft professionals is
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!