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A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water,

Scorpion4ik [409]

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

4 0

3 years ago

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

kondaur [170]

Answer:

Detailed solution is given in the attached diagram

7 0

3 years ago

The increasing interconnections of peoples and countries around the world is known as ________.

Irina18 [472]

I think it’s rationalization.
Hope this helps

4 0

3 years ago

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A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse

d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

$T = T_m + (T_0-T_m)e^{kt}$

where,

$T_0$ = initial temp

$T_m$ = temp of room

$T$ = temp after t hours

$k$ = how fast the temp is changing

$t$ = time (hours)

$T_0 = 31$ because the body was initlally 31ºC when the police found it

$T_m = 22$ because that was the room temp

$T = 30$ because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

$k=???$ we don't know k so we must solve for this

rearrange the equation to solve for k

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k$

plug in the numbers to solve for k

$k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}$

$k = \frac{ln(\frac{30 - 22}{31-22})}{1}$

$k=ln(\frac{8}{9})$

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

$T_0 = 37$ because the body was 37ºC right after being killed

$T_m = 22$ because that was the room temp

$T = 31$ because the body temp when the police found it

$k=ln(\frac{8}{9})$ we solved this earlier

t = ??? we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t$

plug in the values

$t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}$

$t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}$

$t=\frac{ln(3/5)}{ln(8/9)}$

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0

3 years ago

Relay contacts that are defined as being normally open (n.o.) have contacts that are:_____.

Reptile [31]

Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if the relay coil is known to have de-energized.

<h3>What is meant by normally open contacts?</h3>

Normally open (NO) are known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.

Note that Normally closed (NO) contacts are said to be closed only if the coil is said to be de-energized and open only if the coil is said to carry current or is known to have energized.

The role of relay contact is wide. The Relays are tools that are often used in the work of switching of control circuits and it is one that a person cannot used for power switching that has relatively bigger ampacity.

Therefore, Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if the relay coil is known to have de-energized.

Learn more about Relay contacts from

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8 0

1 year ago

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