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coldgirl [10]
3 years ago
5

Question 8 (1 point)

Engineering
1 answer:
Roman55 [17]3 years ago
7 0

Answer:

I believe it is B : Load Index. Please tell me if I'm wrong.

Explanation:

I'm taking Tech/Engineering and my dad works on cars. I don't know too much about tires, but I know a little. If you need help with engineering, Tell me!

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In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.
egoroff_w [7]

Answer:

a. True

Explanation:

A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.

Thus the answer is true.

A system is casual when the output of the system at any time depends on the input only at the present time and in the past.

A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.

A system is linear when it satisfies the additivity and the homogeneity properties.

A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.

Thus a system can be time invariant, memoryless, linear, casual and stable.

4 0
3 years ago
A man releases a stone (at rest, Vo=0) from the top of a tower. During the last second of its travel, the stone falls through a
RideAnS [48]

Answer:

Height of tower equals 122.5 meters.

Explanation:

Since the height of the tower is 'H' the total time of fall of stone 't' is calculated using second equation of kinematics as

Since the distance covered in last 1 second is \frac{9H}{25} and the total distance covered in 't' seconds is 'H' thus the distance covered in the first (t-1) seconds of the motion equals

S_{t-1}=S_{t}-S_{last}\\\\S_{t-1}=H-\frac{9H}{25}=\frac{16H}{25}

Now by second equation of kinematics we have

S=ut+\frac{1}{2}gt^{2}\\\\S=\frac{1}{2}gt^{2}(\because u=0)

Thus we have

\frac{16H}{25}=\frac{1}{2}g(t-1)^{2}.............(i)\\\\H=\frac{1}{2}gt^{2}..............(ii)

Dividing i by ii we get

\frac{16}{25}=\frac{(t-1)^{2}}{t^2}\\\\\therefore \frac{t-1}{t}=\frac{4}{5}\\\\\therefore t=5secs

Thus from equation ii we obtain 'H' as

H=\frac{1}{2}\times 9.8\times 5^{2}=122.5meters

6 0
3 years ago
A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of 10 m ´ 8 m, an effective filtration rate of 7.70 m/h.
galben [10]

Answer:

Explanation:

given data

loading rate = 8.00 m/h

filtration rate = 7.70 m/h

dimensions = 10 m × 8 m

filter cycle duration = 52 h

time = 20 min

to find out

flow rate  and  volume of water is used for back washing plus rinsing the filter  

solution  

we consider here production efficiency is 96%

so here flow rate will be  

flow rate = area × rate of filtration  

flow rate = 10 × 8 × 7.7  

flow rate = 616 m³/h

and  

we know back washing generally 3 to 5 % of total volume of water per cycle so  

volume of water is = 616 × 52

volume of water is  32032 m³

and  

volume of water of back washing is = 4% of 32032  

volume of water of back washing is 1281.2 m³

8 0
3 years ago
Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance
Svetradugi [14.3K]

Answer:

specific\ volume=0.00097\ m^3/kg

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 kg/m^3

Density of sea water = 1025  kg/m^3

We know that

Density=\dfrac{mass}{Volume}   ---1

Specific volume

specific\ volume=\dfrac{Volume}{mass}    ---2

From equation 1 and 2

We can say that

specific\ volume=\dfrac{1}{density}\ m^3/kg

specific\ volume=\dfrac{1}{1025}\ m^3/kg

specific\ volume=0.00097\ m^3/kg

6 0
3 years ago
Answer the following questions briefly<br> 1.Write the definitions of *Comminution and *Calcination
VikaD [51]

Answer:

Comminution is defined as the reduction of solid materials from one average particle size to a smaller average particle size through various methods which include crushing, grinding, cutting, vibrating etc,

Calcination is also defined as a thermal treatment process in the absence or limited supply of air or oxygen applied to ores and other solid materials to bring about a thermal(heat) decomposition.

8 0
3 years ago
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