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3 years ago

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A box-shaped metal can has dimensions 5 in. by 19 in. by 4 in. high. All of the air inside the can is removed with a vacuum pump

. Assuming normal atmospheric pressure outside the can, find the total force (in lb) on one of the 5-by-19 in. sides. (Enter the magnitude.)

1 answer:

GuDViN [60]3 years ago

5 0

Answer:

The force is $F = 1397 lb$

Explanation:

From the question we are told that

The length of the box is $l = 19 \ in$

The width of the box is $w = 5 \ in$

The height is $h = 4\ in$

The pressure experience on one of the sides is mathematically represented as

$p = \frac{F}{A}$

Where A is the area of the box which is mathematically evaluated as

$A = l * w$

substituting values

$A = 5 *19$

$A = 95 \ in^2$

This pressure is equivalent to the atmospheric pressure which has a constant value of $p = 14.7 pi$

This implies that

$14.7 = \frac{F}{95}$

=> $F = 14.7 *95$

=> $F = 1397 lb$

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Answer:

$h=25.52\times10^6 m$

Explanation:

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Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

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Solution:

Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface

$-G M m / ( R + h )=- G M m / R + (1/2) m v^2$

$- G M / ( R + h ) = - G M / R + (1/2) v^2$

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= $( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2$

= $-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

$R+h=31.92\times10^{6}$

$h=31.92\times10^{6}-6.4\times10^6$

$h=25.52\times10^6 m$

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3 years ago

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.

katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

M v + m v' = 0

5000 x v - 10 x 4000 = 0

5000 v = 40000

v = 8 m/s

speed of rocket = 8 m/s

now,

we know

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$F = \dfrac{5000(8-0)}{10}$

F = 4000 N

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Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

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A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency

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Answer:

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v = speed of the wave = ?

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v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

$\mu = \frac{m}{L}$

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Answer:

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velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

$\omega = \sqrt{\dfrac{k}{m}}$

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$a_x = \dfrac{k}{m}x$

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