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3 years ago

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In a hydraulic system, a 100.-newton force is applied to a small piston with an area of 0.0020 m2. What pressure, in pascals, wi

ll be transmitted in the hydraulic system?

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

50000 Pascals

Explanation:

This is a simple equation where Pascals=Force/Area. 100/0.002 = 50000.

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What is the difference between pump and turbine? Write the first law of thermodynamics for both (pump & turbine)?

tiny-mole [99]

Answer:

Pumps converts mechanical energy into hydraulic energy while turbines convert hydraulic energy into mechanical energy.

Explanation:

The machines which converts and transfers mechanical energy in the form of torque on the shaft into hydraulic energy in the form of water under pressure are called pumps whereas those machines which converts water pressure or hydraulic energy into mechanical energy that is further converted into electrical energy are called turbines.

The pump impeller rotates in the opposite direction to the turbine runner.

A turbine delivers work as output whereas a pump consumes work.

First law of thermodynamics for a pump :

W = ( H₁-H₂) +Q , where H₁ > H₂

First law of thermodynamics for a turbines :

W = ( H₂-H₁) +Q , where H₁ < H₂

8 0

3 years ago

A motor car is travelling at 144km/h in a 90km/h speed zone .The driver suddenly sees speed camera 80m ahead before the camera c

Kruka [31]

The deceleration required to comply with the speed limit before being caught by the camera is 6.094 meters per square second.

Let assume that the motor car <em>decelerates</em> at <em>constant</em> rate. Given the <em>initial</em> and <em>final</em> speeds ( $v_{o}$ , $v$ ), in meters per second, and <em>travelled</em> distance ( $s$ ), in meters, the deceleration ( $a$ ), in meters per square second, is determined by this formula:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot s}$

If we know that $v_{o} = 25\,\frac{m}{s}$ , $v = 40\,\frac{m}{s}$ and $s = 80\,m$ , then the deceleration required to comply with the speed limit is:

$a = \frac{\left(25\,\frac{m}{s} \right)^{2}-\left(40\,\frac{m}{s} \right)^{2}}{2\cdot (80\,m)}$

$a = -6.094\,\frac{m}{s^{2}}$

The deceleration required to comply with the speed limit before being caught by the camera is 6.094 meters per square second.

We kindly invite to check this question on uniform accelerated motion: brainly.com/question/12920060

4 0

3 years ago

Given a dictionary d and a list lst, remove all elements from the dictionary whose key is an element of lst. For example, given

NikAS [45]

Answer:

d = {1:2, 3:4, 5:6, 7:8}

list = [1, 7]

not_found = set()

for x in list:

if x not in d.keys():

not_found.add(x)

else:

del d[x]

print(d)

print(not_found)

6 0

3 years ago

for the given sand, the maximum and minimum dry unit weights are 108 lb/ft3 and 92 lb/ft3, respectively. given that Gs=2.65, det

raketka [301]

Answer:

The moist unit weight of compaction = 109.05 lb/ft3

Explanation:

In order to determine the moist unit weight, the dry unit weight has to be evaluated first. If Y is the moist unit weight, then:

Y = Yd (1 + m)

Where:

Yd = dry unit weight

m = moisture content of soil = 8% = 0.08

But the dry unit weight is unknown. In order to calculate the dry unit weight, we will make use of the formula for relative density R;

R = [(Yd — Ydmin) ÷ (Ydmax — Ydmin)] × [Ydmax ÷ Yd]

Where:

R = relative density = 60% = 0.6

Yd = dry unit weight

Ydmin = minimum dry weight = 92 lb/ft3

Ydmax = maximum dry weight = 108 lb/ft3

Therefore R = 0.6 = [(Yd — 92) ÷ (108 — 92)] × [108/Yd]

0.6 = [(Yd — 92)/16] × [108/Yd], or

0.6 = (0.0625Yd — 5.75) × [108/Yd]

0.6Yd = 6.75Yd — 621

6.75Yd — 0.6 Yd = 621

6.15Yd = 621

And Yd = 100.98 lb/ft3 = dry unit weight

But we are asked to find the moist unit weight = Y = Yd (1 + m)

where Yd = dry unit weight and m = moisture content of soil = 8% = 0.08

Therefore, Y = 100.98 (1 + 0.08) = 109.05 lb/ft3.

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3 years ago

ME<br> And what you know about love

antoniya [11.8K]

Answer:

Everything I got what you need

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3 years ago