3 years ago

PLZ PLZ PLZ PLZ PLZ PLZ PLZ HELPPPPPPPPPPPPPPPPP

Physics

2 answers:

pshichka [43]3 years ago

4 0

C is the answer I think

Dimas [21]3 years ago

3 0

Answer:

endoplasmic reticulum

You might be interested in

What is the push and pull when two kids are on a seesaw and one of them is in the air? What is causing a change in SPEED, or DIR

balandron [24]

That doesn't make sense I don't think it cause eaither

3 0

3 years ago

As much as you love high school, especially your Physics class, you're super excited for 2026 when the US and Mexico will host t

Veseljchak [2.6K]

3.63 years is the right one

3 0

3 years ago

Read 2 more answers

What process change earths surface during Precambrian time?

andreyandreev [35.5K]

The three eons of precambrian time i.e hadean eon ,Archean eon and proyerozoic eon change earth's surface during Precambrian time.

4 0

3 years ago

A scientist agrees with the scientific consensus that dinosaurs have scaly skin. Then he discovers a new fossil that indicates a

patriot [66]

The scientist is biased

3 0

3 years ago

Read 2 more answers

A centrifugal pump rotates at n˙ = 740 rpm. Suppose the pump has some swirl at the inlet (α1 = 10°) and exits at an angle of 35°

Blizzard [7]

Answer:

Net head = 380cm

bhp = 17.710kW

Explanation:

Angular velocity of centrifugal pump:

$w=\frac{2\pi n}{60}=\frac{2\pi (750)}{60}=78.54\frac{rad}{s}$

Normal velocity component at outlet of pump:

$V_{2,n}=\frac{V}{2\pi r_{2}b_{2}}=\frac{0.573}{2\pi (0.24)(0.162)}}=2.346\frac{m}{s}$

Tangential velocity component at exit of the pump:

$V_{2,t}=v_{2,n}tan\alpha _{2}=(2.346)tan(35)=1.643\frac{m}{s}$

Normal velocity component at inlet of pump:

$V_{1,n}=\frac{V}{2\pi r_{1} b_{1}}=\frac{0.573}{2\pi (0.12)(0.18)}=4.22\frac{m}{s}$

Tangential velocity component at inlet of the pump:

$V_{1,t}=v_{1,n}tan\alpha _{1}=(4.22)tan(0)=0\frac{m}{s}$

Equivalent head in centimetre of water column:

$H_{water}=H(\frac{rho_{air}}{rho_{water}})\\\\H_{water}=(\frac{w}{g} )(r_{2}V_{2,t}-r_{1}V_{1,t})(\frac{rho_{air}}{rho_{water}}) \\\\H_{water}=(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0)})(\frac{1.2}{998})=38m=380cm$

Break horse power:

$bhp=rho_{water} gHV=rho_{water} g[(\frac{w}{g})(r_{2}V_{2,t}-r_{1}V_{1,t})]V\\\\bhp=(998)(9.81)[(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0))](0.573)=17710W=17.710kW$

8 0

3 years ago