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Marizza181 [45]
3 years ago
10

At what voltage would a () capacitor have the minimum energy to raise by ignoring all losses in the system? If needed, you may a

ssume a gravitational acceleration of 9.80 m/s2.

Engineering
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

V = 280.15 V

Explanation:

" The complete question is attached with figure"

Given:

- The capacitance of the capacitor C = 10 nF

- The amount of mass attached to motor m = 4 grams

- The amount of distance it is to be lifted h = 1 cm

- Ignore all other losses in the system

Find:

- The voltage required to lift the mass m through distance h?

Solution:

- The conservation of energy for the entire system is written as:

                            Work_gravity = U_c

Where,

           Work_gravity: Work done by gravity on mass m

           U_c: The amount of energy stored in a capacitor

                             m*g*h = 0.5*C*V^2

                             V^2 = 2*m*g*h / C

                             V = sqrt ( 2*m*g*h / C )

Plug in the values:

                             V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )

                             V = sqrt ( 78,480)

                             V = 28.15 V

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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
What does the word “robot” mean? A.Clone B. Athlete C. Servant D. Actor
hram777 [196]

Answer:

a. clone

Explanation:

4 0
3 years ago
If you were driving the blue Prius in the situation pictured above, explain why the red Mustang should be given right-of-way at
dexar [7]

Answer:

The blue Prius, because the Mustang arrived almost in the same time. And when you arrive in an intersection at the same time of other vehicle you need yield for the car on your right if the car is on your left you have the right of way.

Explanation:

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3 years ago
Tires can be recycled instead of thrown out.<br> True<br> False
Arisa [49]

Answer:

True :)

Explanation:

You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.

Hope it helped have a nice day! :)

8 0
2 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

4 0
3 years ago
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