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3 years ago

11

Who has good grades in school like A's B's dm me

2 answers:

serious [3.7K]3 years ago

8 0

Why do you need help with school work

siniylev [52]3 years ago

4 0

Answer:

I do

Explanation:

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A circle has an area of 153.86 in squared what’s the radius of the circle

sweet [91]

Answer: 49

Explanation: So, how you figure this out is by taking the area of the circle (153.86) and dividing it by pi (3.14) which equals 49 which is your answer! So, the <u>radius of the circle is </u><u>49</u>.

If you need any further instruction or help let me know!

6 0

3 years ago

A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the lower end of a slender 23-lb bar of length L

Sergeeva-Olga [200]

Answer: hello attached below is the missing image the slender weight is different from what is in the question here so I worked with 23-Ib as requested in the question

answer

≈ 12.17 Rad/sec

Explanation:

weight of bullet ( Wb ) = 0.08 Ib

horizontal velocity = 1800 ft/s

Slender(Wr) = 23-Ib bar with

length ( L ) = 30

h = 12 inches

Vro = 0

<u>Calculate the angular velocity of the bar immediately after the bullet becomes embedded </u>

attached below is a detailed solution

6.708 = ( 0.05011 + 0.5011 ) w'

w' = 6.708 / 0.55121 ≈ 12.17 Rad/sec

6 0

3 years ago

Oil system cleaning products should not use solvents problem

pantera1 [17]

Sorry I don’t know this one out I need points sorryyyy

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3 years ago

From an email message sent to all employees of a law firm advising them to be alert for a new virus: This virus has been known t

Fudgin [204]

Answer:true

Explanation: I’m fram wokonda

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3 years ago

A horizontal water jet at 70 °F issues from a circular orifice in a large tank. The jet strikes a vertical plate that is normal

qaws [65]

Answer:

The diameter of the jet just down the orifice is 0.326 ft

Explanation:

Temperature of the horizontal water = 70⁰C

Force, F = 600 lbf

$P_{A} = 25 psig$

Force, F = $\frac{mv^{2} }{r}$

Mass, m = $\rho V$ , where $\rho =$ density of water = 62.4 lbs/ft³

V = volume

F = $\frac{\rho Vv^{2} }{r}$

Area, A = V/r

$F = \rho A v^{2}$ ...........(1)

Applying Bernoulli's equation between point A and B on the orifice

$\frac{P_{A} }{\rho} + \frac{V_{A} ^{2} }{2}+ gh_{A} = \frac{P_{B} }{\rho} + \frac{V_{B} ^{2} }{2}+ gh_{B}$ ..................(2)

$h_{A} = h_{B}$

At point A, the initial rest position, $v_{A} = 0$

At the orifice, $P_{B} = 0$

Making the appropriate substitution, equation (2) becomes

$\frac{P_{A} }{\rho} = \frac{v_{B} ^{2} }{2}$

$P_{A} = 25 psig\\P_{A} = 25 * 144 psf\\P_{A} = 3600 psf$

$\rho = 1.94 slug/ft^{3}$

$\frac{3600 }{1.94} = \frac{v_{B} ^{2} }{2}\\v_{B} ^{2} = 3711.34\\v_{B} =\sqrt{3711.34} \\v_{B} = 60.92 ft/s$

Put the value of $v_{B}$

$600 = 1.94* A 60.92^{2}$

600 = 4320A

A = 600/7200

A = 0.0833 ft²

Area, $A = \frac{\pi d^{2}} {4}$

$0.0833 = \frac{\pi d^{2}} {4}\\d^{2} = (4*0.0833)/\pi \\d^{2} = 0.106\\d = 0.326 ft$

8 0

4 years ago