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3 years ago

PLEASE HELP, PLEASE A CORRECT ANSWER!

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Answer: I like your profile picture

Explanation:

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Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.

Sergeu [11.5K]

The work done to pull the sled up to the hill is given by
$W=Fd$
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is $W=1900 J$ and the distance through which the force is applied is $d=400 m$ , so we can calculate the average force by re-arranging the previous equation and by using these data:
$F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N$

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4 years ago

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iragen [17]

You’re correct, it’s gravity

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3 years ago

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2.<br> ___ ___ is the number of wavelengths that pass a given point/second.

atroni [7]

Answer:

frequency tell me if im right

Explanation:

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4 years ago

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:

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

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3 years ago

Does a North magnet and south magnet repel or attract?

NNADVOKAT [17]

Answer:

A north magnet attracts a south magnet

Explanation:

The opposite polar magnetic field will attract each other while the same polar magnetic fields will repel each other.

5 0

3 years ago

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PLEASE HELP, PLEASE A CORRECT ANSWER!​

PLEASE HELP, PLEASE A CORRECT ANSWER!