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2 years ago

What is Not considered as metric system ? a. Length b. Mass c. Time d. Volume e. Temperature

Engineering

2 answers:

vampirchik [111]2 years ago

7 0

Answer:

D.) Volume

Explanation:

borishaifa [10]2 years ago

5 0

Answer:

c. Time

Explanation:

The metric system measures mass in grams

The metric system measures mass in grams or kilograms, distance in meters or kilometers, and volume in liters. It measures temperature in Kelvin or Celsius degrees instead of the Fahrenheit degrees used in the imperial system.

Metric is based on increments of the number 10; english is a linear measurement based on inches, feet and yards.

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Que rol tiene el ecosistema el patos

zhenek [66]

Answer:

Investigué un poco y descubrí algunas cosas sorprendentes sobre los patos.Espero que los encuentres interesantes:Las aves acuáticas como los patos pueden mantener la diversidad de otros organismos, controlar las plagas, ser bioindicadores efectivos de las condiciones ecológicas y actuar como centinelas de posibles brotes de enfermedades. También proporcionan importantes aprovisionamientos (carne, plumas, huevos, etc.) y servicios culturales a las sociedades indígenas y occidentalizadas.La ubicuidad a menudo ridiculizada de los patos los convierte en vehículos ideales para transportar semillas de un lugar a otro, y eso significa humedales y biodiversidad más saludables para el beneficio de todas las aves y la vida silvestre.Sabías ? :Los patos ayudan al medio ambiente porque limpian los químicos del agua para que sus amigos y otros animales no mueran y se enfermen gravemente.

Explanation:

3 0

2 years ago

A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret

andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

<u>Explanation:</u>

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = $\frac{8000 X 10^-^3}{0.7}$

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = $\frac{0.002}{5}$

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

$E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi$

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0

2 years ago

A teenager was pulling a prank and placed a large stuffed penguin in the middle of a roadway. A driver is traveling on this leve

Anvisha [2.4K]

Whats the question????

3 0

3 years ago

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0

2 years ago

Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago