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kicyunya [14]
3 years ago
9

With increases in magnification, which of the following occur? a. The field of view decreases. b. The ambient illumination decre

ases. c. The larger parts can be measured. d. The eyepiece must be raised.
Engineering
1 answer:
Irina-Kira [14]3 years ago
7 0

By increasing magnification you decrease the field of view.

The answer is A.

Hope this helps.

r3t40

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3.94 x 105) + (2.04 x 105)
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627.9 is the answer
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2 years ago
Who will win the Copa America 2021 and Euros 2021?
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Brazil will win copa America
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3 years ago
Which of the following headlines about the French Revolution is not true?
Sever21 [200]

Answer:

d) "Napoleon Declares Himself Holy Roman Emperor"

Explanation:

The French Revolution is defined as a period of the major social upheaval which began in the year 1787 and lasted till year 1799. This revolution completely redefined the the very nature of the political power in France. and also the relationship between the rulers of France and the people they governed.

The 1789 Estates-General was the 1st meeting since year 1614 of the French Estates-General. It is a general assembly which represents the French estates of  realm.

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Maximilien Robespierre was considered to be one of the most influential figure and most important statesman during the French Revolution.

Thus all the options (a),(b) and (c) are headlines about the French Revolution, except option (d).

3 0
3 years ago
Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und
sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of  each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is v  = cd\sqrt{\frac{2*P}{p}}

v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec

injected fuel volume  (V) =Area of given  Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

SoV = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec

putting these value in volume equation and solve for Discharge 8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times  375 \times  \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}

Q = 8.845 DEGREE

4 0
2 years ago
Suppose that units of force, length, and time are chosen such that the density of water and the acceleration of gravity are both
k0ka [10]

Answer: So a 1 pound = 0.453592kg

the density of the water is D= 997 kg/m^{3} so we need to use replace the kg with pounds with 2.20462 pounds = 1kg.

then D = 2.20462*997 pounds/m^{3}

and we want to make this equal to one changing the metters for other thing.

we need to replace x^{3} with 2.020462*997 meters qube.

then our new metters variable x = 12.6meters.

now g = 9.8 meters for second square.

in our new lenght variable x.

g = 12.6*9.8 x for second square.

so if y is our new time variable

y^{2} = 12.6*9.8 seconds square. so y = 11.1seconds

so the units of length are x = 12.6 meters and for time y = 11.1 seconds.

3 0
2 years ago
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