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gladu [14]
3 years ago
9

What causes a boat to float? My questions are super easy. So you can follow me if you like easy question.

Engineering
2 answers:
marin [14]3 years ago
6 0
Boats float because the gravity is acting down on it and the buoyant force is acting up on the ship.
amid [387]3 years ago
4 0

Answer:

i think it A

Explanation:

You might be interested in
A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem
lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is \Delta h=9.975\,m

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature.  As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2}  \Delta v^2 =0

(\rho stands here for density, h for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

\frac{\Delta P}{\rho}+g\, \Delta h  =0

\Delta P= -g\, \rho\, \Delta h

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\

We insert that into our last equation and get:

\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0
3 years ago
If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical
vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 ×10^{6} N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 ×10^{9} N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}    .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm  = 0.3 ×10^{-3} m

so now put value in equation 1 we get

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}

( σc ) = \sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}

( σc ) = 27.396615 ×10^{6} N/m²

so critical stress required for the propagation is 27.396615 ×10^{6} N/m²

6 0
3 years ago
Water (density p-1000 is discharging from through a hole at the bottom of a graduated 71 cylinder. The mass flow rate exiting th
alexira [117]

Answer:

Please see attachment

Explanation:

Please see attachment

4 0
3 years ago
What is the shape of the output signal on a rigexpert analyzer?
Gekata [30.6K]

Answer:

Output signal shape: square, from 0.1 to 230 MHz. Output power: -10 dBm (at a load of 50 Ohms).

Explanation:

8 0
2 years ago
Read 2 more answers
LC3 Programming ProblemUse .BLKW to set up an array of 10 values, starting at memory location x4000, as in lab 4.Now programmati
irga5000 [103]

Answer:

Check the explanation

Explanation:

Code

.ORIG x4000

;load index

LD R1, IND

;increment R1

ADD R1, R1, #1

;store it in ind

ST R1, IND

;Loop to fill the remaining array

TEST LD R1, IND

;load 10

LD R2, NUM

;find tw0\'s complement

NOT R2, R2

ADD R2, R2, #1

;(IND-NUM)

ADD R1, R1, R2

;check (IND-NUM)>=0

BRzp GETELEM

;Get array base

LEA R0, ARRAY

;load index

LD R1, IND

;increment index

ADD R0, R0, R1

;store value in array

STR R1, R0,#0

;increment part

INCR

;Increment index

ADD R1, R1, #1

;store it in index

ST R1, IND

;go to test

BR TEST

;get the 6 in R2

;load base address

GETELEM LEA R0, ARRAY

;Set R1=0

AND R1, R1,#0

;Add R1 with 6

ADD R1, R1, #6

;Get the address

ADD R0, R0, R1

;Load the 6th element into R2

LDR R2, R0,#0

;Display array contents

PRINT

;set R1 = 0

AND R1, R1, #0

;Loop

;Get index

TOP ST R1, IND

;Load num

LD R3,NUM

;Find 2\'s complement

NOT R3, R3

ADD R3, R3,#1

;Find (IND-NUM)

ADD R1, R1,R3

;repeat until (IND-NUM)>=0

BRzp DONE

;load array address

LEA R0, ARRAY

;load index

LD R1, IND

;find address

ADD R3, R0, R1

;load value

LDR R1, R3,#0

;load 0x0030

LD R3, HEX

;convert value to hexadecimal

ADD R0, R1, R3

;display number

OUT

;GEt index

LD R1, IND

;increment index

ADD R1, R1, #1

;go to top

BR TOP

;stop

DONE HALT

;declaring variables

;set limit

NUM .FILL 10

;create array

ARRAY .BLKW 10 #0

;variable for index

IND .FILL 0

;hexadecimal value

HEX .FILL x0030

;stop

.END

7 0
3 years ago
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