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3 years ago

What is the best reason a scientist might use a model to represent the solar system?

Engineering

1 answer:

uranmaximum [27]3 years ago

3 0

So they can grasp a better understanding! good luck

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HELP!

olya-2409 [2.1K]

The thickness is thick

5 0

3 years ago

An air conditioning system is to be filled from a rigid container that initially contains 5 kg of saturated liquid at 24° Celsiu

gtnhenbr [62]

And air-conditioning system is to be filled for my ridge the containerBut that internally contains 5 kgDetermine the final quality of the arm 134

7 0

2 years ago

According to fire regulations in a town, the pressure drop in a commercial steel, horizontal pipe must not exceed 2.0 psi per 25

bonufazy [111]

Answer:

6.37 inch

Explanation:

Thinking process:

We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.

To determine the pressure drop in the pipe:

Using the Bernoulli equation for mass conservation:

$\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1} = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}$

thus

$\frac{P1-P2}{\rho } = f\frac{l}{D} \frac{v^{2} }{2g}$

The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.

Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F

from the tables

Re= 2.01 × 10⁵

Hence, f = 0.018

Therefore, pressure drop, (P1-P2)/p = 2.70 ft

This occurs at ae presure change of 1.17 psi

Correlating with the chart, we find that the diameter will be D= 0.513

= <u>6.37 in Ans</u>

7 0

3 years ago

Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1

tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R

where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0

3 years ago

What happens when force is placed on a square/rectangle?

mart [117]

im not sure i need to see a photo and also is this science

7 0

3 years ago

Read 2 more answers