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blondinia [14]
3 years ago
10

Determine the maximum height (in inches) that a lift pump can raise water (0.9971 g/ml) from a well at normal atmospheric pressu

re. Also present the cancellation of units to arrive at the final unit? Pls help. with solution​
Engineering
1 answer:
solniwko [45]3 years ago
7 0

Answer:Height = 30 inches​

Mass of water / Volume of water x Gravity constant for Earth x Density of Water. This cancels out to give us mass/volume, which is then cancelling mass, giving us the final unit of length/area (in this case inches).​

Explanation:

Height = 30 inches.

Units cancelled:

30/30 x 9.8 x 0.9971 = 30/39.97

The final units are in inches!​

2. Determine the weight (in pounds) that an ice cube at 32°F will displace when placed in water at 75°F.

Weight of ice cube / Volume of water x Gravity constant for Earth x Density of Ice = Weight displaced by ice cube in pounds​

Weight of ice cube / Volume of water x Gravity constant for Earth x Density of Ice = 0.5/75 x 32 x 0.92 = 0.5/22.67

The final units are in pounds!​

3. Determine the mass (in grams) of an ice cube at 32°F when placed in water at 75°F (assume no change in volume).

Mass of Ice Cube / Volume of Water x Gravity constant for Earth x Density of Ice = Mass of ice cube in grams

Mass of Ice Cube / Volume of Water x Gravity constant for Earth x Density of Ice = 0.5/75 x 32 x 0.92 = 15/22.67

The final units are in grams!​

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cupoosta [38]

Answer:

Depends on the battery and the current type.

Is it AC or DC?

Explanation:

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8 0
3 years ago
The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current
Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}

\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

If i = 2A and g = 10 cm

\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97

f_{m}=-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N

b) Using the coenergy of the system:

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2}  }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N

8 0
3 years ago
Select the correct answer. which process involves creating a product by heating metals and changing their shape through the appl
Paraphin [41]

Answer:

I would say that it is forming.

Explanation:

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4 0
3 years ago
A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.
Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, \mu = 2.501 V

standard deviation, \sigma = 0.113 V

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, P(X\leq 2.70):

z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761

Now, from the Probability table of standard normal distribution:

P(z\leq 1.761) = 0.9608

Now,

P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%

Now, for the expected no. of readings greater than 2.70 V:

P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}

No. of readings expected to be greater than 2.70 V = P(X\geq 2.70)\times Total\ no.\ of\ readings

No. of readings expected to be greater than 2.70 V = 0.0392\times 60 = 2.352 ≈ 2

7 0
3 years ago
Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e
Lady_Fox [76]

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

a)

from the voltage divide rule

V_{in} = V_0 * \frac{R_2}{R_2 + R_f}

\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20

\frac{R_f}{R_2} = 19

Select the nearest answer

b)

obtained gain = \frac{V_0}{V_{in}} = 1 + \frac{36}{2} = 19

Expected gain = \frac{V_0}{V_{in}} = 20

∴ error = |\frac{19 - 20}{20}| × 100

            = 1/20 × 100            

            = 5%

6 0
3 years ago
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