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BlackZzzverrR [31]
3 years ago
12

For installation with a 25-kVA, 3-phase transformer, a 440-volt primary, and a 120-volt secondary. Calculate the maximum overcur

rent protection value (not the fuse or circuit breaker rating) using the protection method for primary-only protection. (Round the FINAL answer to two decimal places.)
Physics
1 answer:
andre [41]3 years ago
4 0

Answer:

41.053 A

Explanation:

given,

three phase kVA = 25-kVA

voltage = 440 Volt

current = ?

To determine three phase kVA when volts and amperes are know

   three phase kVA = 1.73 x V x I

    I = \dfrac{three\ phase\ kVA}{1.73 \times V}

    I = \dfrac{25000}{1.73 \times 440}

           I = 32.84 A

the maximum overcurrent protection value is equal to

            = 125 % of I

            = 1.25 x 32.84

            = 41.053 A

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The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le
Vinil7 [7]

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

3 0
3 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
  • This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
5 0
2 years ago
4. What is the momentum of a 70 kg object traveling at 20 m/s?
Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

6 0
3 years ago
If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52
castortr0y [4]

Answer:

A.\hspace{3}n=1.52

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

n=\frac{c}{v}   (1)

Where:

c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\

v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}

Finally, replacing the data in (1):

n=\frac{300,000}{196339.48} =1.527965746\approx1.52

3 0
3 years ago
What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?​
aleksley [76]

Answer:

250,000

Explanation:

<h2> </h2>

<h2>formula = ( F=ma </h2>

  • F=1500N
  • a=6m/s^2
  • F= ma
  • m=?
  • 1500/6 = m
  • m=250 kg
  • 1kg =1000gm so 250kg =250,000gm
  • m =250×10^3 gm
5 0
3 years ago
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