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3 years ago

12

What is the relationship between the line current i l and the phase current i1 in a balanced delta system?

1 answer:

Natasha_Volkova [10]3 years ago

6 0

<span>In a star connection the line current is equal to the phase current . While in Delta connection the line current is divided by under root 3 to get the phase current.</span>

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A rectangular loop of copper wire of resistance r has width a and length

sergij07 [2.7K]

Answer:

Explanation:

b = b₀ cos ω t

When t = 0 , magnetic field will be b₀ and positive or directed into the page . This is the maximum value of magnetic field. As times goes ahead , magnetic field decreases so magnetic flux decreases . The induced emf or current will be such that it will opposes this reduction of magnetic field. Hence , current in clockwise direction will be generated in the coil which will generate magnetic flux into the paper.

In this way current will be induced clockwise.

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2 years ago

A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i

Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

$m_1v_1+m_2v_2=(m_1+m_2)V$

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

$v_2=-\dfrac{m_1v_1}{m_2}$

$v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}$

$v_2=-1.47\ m/s$

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0

3 years ago

A soccer ball is kicked At 8m/s. It lands on the ground after being in the air for .85 seconds. At what angle is it kicked?

SpyIntel [72]

If ball remains in air for total time T = 0.85 s

this is also known as time of flight

In order to find the time of flight we can use kinematics

$\Delta Y = v_y*t + \frac{1}{2}at^2$

so for complete motion its displacement in y direction will be zero

$0 = v_y* 0.85 + \frac{1}{2}(-9.8)(0.85^2)$

$0 = v_y*0.85 - 3.54$

$v_y = 4.165 m/s$

now we know that net velocity of the ball is 8 m/s

while is y direction component we got is vy = 4.165 m/s

now by component method we can say

$v_y = v sin\theta$

$4.165 = 8 sin\theta$

$\theta = sin^{-1}\frac{4.165}{8}$

$\theta = 31.4^0$

so it is projected at an angle of 31.4 degree above horizontal

8 0

3 years ago

A car of mass 487 kg travels around a flat, circular race track of radius 53.3 m. The coefficient of static friction between the

aleksklad [387]

Answer:

9.96 m/s

Explanation:

mass of car, m = 487 kg

radius of track, R = 53.3 m

coefficient of static friction, μ = 0.19

acceleration due to gravity, g = 9.8 m/s^2

let v be the maximum speed so that the car can go without flying off the track.

The formula for the maximum speed is given by

$v_{max}=\sqrt{\mu Rg}$

$v_{max}=\sqrt{0.19\times53.3\times9.8$

vmax = 9.96 m/s

8 0

3 years ago

A 11.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1985 J by the time the rock

nirvana33 [79]

Answer:

h = 18.41 m

Explanation:

Given that,

Mass of a test rocket, m = 11 kg

Its fuel gives it a kinetic energy of 1985 J by the time the rocket engine burns all of the fuel.

According to the law of conservation of energy,

PE = KE = mgh

h is height will the rocket rise

$h=\dfrac{E}{mg}\\\\h=\dfrac{1985 }{11\times 9.8}\\\\h=18.41\ m$

So, the rocket will rise to a height of 18.41 m.

5 0

2 years ago