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damaskus [11]
3 years ago
14

What should the resistance value be on a size 5 motor starter coil

Engineering
1 answer:
Katen [24]3 years ago
4 0

Answer:

1000 Mega Ohms.

Explanation:

One mega Ohm is equal to 1,000,000 ohms, which is the resistance between two points of a conductor with one ampere of current at one volt. The mega Ohm is a multiple of the Ohm, which is the SI derived unit for electrical resistance. In the metric system, "mega" is the prefix for 10 raised to the power of 6.

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Define waves as it applies to electromagnetic fields
julsineya [31]

Waves in the electric and magnetic fields are known as electromagnetic waves. You must first understand what a field is, which is just a technique of giving each square inch of space a numerical value. You may see that as a temperature field, for instance, when you look at the weather predictions and they mention the temperature in several locations. Every location on Earth has a unique temperature that can be quantified. Everywhere on Earth has its own wind velocity, which is another form of field. This field differs somewhat from the temperature field in that the wind velocity has both a direction and a magnitude, whereas the temperature just has a magnitude (how hot it is). A vector is a quantity that has both magnitude and direction, hence a field that contains vectors at every location is referred to as a vector field. Vector fields include the magnetic and electric fields. We may examine what would happen if we placed a charged particle at any given position in space. If the charged particle were to accelerate, we would state that the electric field there is the direction in which the particle is moving. In general, positively charged particles will move in the electric field's direction, whereas negatively charged particles will move in the opposite way. Because it is a vector field, the magnetic field exhibits comparable behavior. We discovered in the 19th century that the same interaction, electromagnetism, really produces both electric and magnetic fields. Like an electromagnet, a changing electric field will produce a magnetic field, and a changing magnetic field will induce an electric field (like in a generator). If your system is configured properly, you may have an electric field that fluctuates, which in turn produces a magnetic field, which in turn induces another electric field, which in turn generates another magnetic field, and so on indefinitely. At the speed of light, this oscillation between a strong magnetic field and strong electric field spreads out indefinitely. In reality, light is an electromagnetic wave—an oscillation in the electromagnetic fields. An electric or magnetic field may exist without a medium since they exist in a vacuum, which implies that waves in these fields don't require a medium like sound to flow through.

5 0
2 years ago
We have a tube with a diameter of 5 inches that is 1 foot long. The tube then reduces the diameter to 3 inches. According to the
Ksju [112]
Can I get a picture? Of the question so I can understand it more?
6 0
3 years ago
Read 2 more answers
A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst
expeople1 [14]

Answer:

s= 20.4 m  

Explanation:

First lets write down equations for each ball:  

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:  

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b  

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s  

now we just have to put that time in any of those equations an get distance from the ground:  

s = 30 + 5*2 -1/2*9.81 *2^2  

s= 20.4 m  

6 0
3 years ago
It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc
ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength (\sigma_{cd})=630\ MPa = 630*10^6\ Pa, Fracture strength

(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa

a) The critical length (L_c) is given by:

L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}}  \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456

6 0
3 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
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