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Makovka662 [10]
3 years ago
12

Annie has collected several items from around her house. She is using these objects to investigate which objects are attracted t

o magnets and which ones are not. Which item should Annie expect to be attracted to her magnet?
A) ice cube

B) metal screw

C) leather glove

D) notebook paper
Engineering
1 answer:
Margarita [4]3 years ago
5 0
The metal screw would be attracted to her magnet since its metal and the others don’t attract to magnets.
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Of the oil I hope this help
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The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo
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Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)

consider the equation of bernoulli at the top of the pool

P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)

where P_1=P_0 atm pressure

At the top of the pool v_1=0m/s, substitute in V_1 in equation (2)

P_0+\rho gh_1 =constant ...(3)

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)

where P_2=P_0 atm pressure and h_2=0m

P_0+\rho v_1^2 =constant ...(6)

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

        P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)    

Hence velocity is v_2=(\sqrt{2gh_1})m/s

d.) consider (7)

v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)

This is the norminal value of velocity  

e.) consider the equation of flow rate interval of v and t

flow(t)=\frac{dv}{dt}(m^3/s) hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2

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Do you think that individuals with the same disabilities have the same needs? If your answer is yes, how? /No, why?​
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Answer:

Explanation:

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