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devlian [24]
3 years ago
7

The multipoint grounded neutral is intended to reduce the _____ neutral voltage drop, assist in clearing _____ line-to-neutral f

aults, and reduce elevated voltage caused by line-to-ground faults.
A. primary / primary
B. secondary / secondary
C. primary / secondary
D. secondary / primary
Physics
1 answer:
Shtirlitz [24]3 years ago
5 0

Answer:

The answer is A, primary/primary.

Refer below for the explanation.

Explanation:

The multipoint grounded neutral is intended to reduce the primary neutral voltage drop, assist in clearing utility line-to-neutral faults, and reduce elevated voltage caused by line-to-ground faults.

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Which tower is more likely to fall?
shepuryov [24]
I say B.  only because its smaller in width and it has a better chance of falling 
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It is winter in Puerto Rico. Compare the air temperatures beachgoers feel near the water
Leviafan [203]
Large amounts of water do have a big impact on the weather: indeed, it takes less energy to warm/cool land than water.
Therefore, places near large amounts of water tend to have smaller differences in temperature between summer and winter than places far from waters.

Hence, during winter in Puerto Rico, alongside the coast, the temperature will be higher than in the innermost parts of the island.
7 0
3 years ago
Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A
Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>

<em> How much work W does the motor do on the platform during this process?  Enter your answer in joules to four significant figures.</em>

The amount of work done by the motor is given by:

W=\Delta K

W= 1/2*I*\omega f^2-1/2*I*\omega o^2

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

\omega f^2=\omega o^2+2*\alpha*\theta

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

\tau=I*\alpha

\alpha=\tau/I     =>     \alpha = 0.5rad/s^2

Now we can calculate the final velocity:

\omega f = 8.68rad/s

Finally, we calculate the total work:

W= 1/2*I*\omega f^2 = 1883.56J

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

3 0
3 years ago
Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement
lana [24]

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

8 0
3 years ago
What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?
vfiekz [6]

Answer: 0.0039\ A

Explanation:

Given

Diameter of the rod d=2.54\ cm

length of rod is l=20\ cm

Resistivity of silicon is \rho=6.4\times 10^2\ \Omega-m

cross-section of the rod A

\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2

Resistance of rod is  R

\Rightarrow R=\dfrac{\rho l}{A}

\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega

Current is given by

\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A

3 0
2 years ago
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