Which of the following would have the strongest electric field? MULTIPLE CHOICE.

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85

kvasek [131]

Answer:

$\mathtt{Q_{sh} = 600.75 \ vars}$

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ = cos⁻¹ (0.85)

θ₂ = 31.788°

However; the initial reactive power $Q_1$ = P×tanθ₁

the initial reactive power $Q_1$ = 1500 × tan(45.573)

the initial reactive power $Q_1$ = 1500 × 1.0202

the initial reactive power $Q_1$ = 1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

$Q_{sh} = P( tan \theta_1 - tan \theta_2)$

$Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)$

$Q_{sh} = 1500( 1.0202 - 0.6197)$

$Q_{sh} = 1500( 0.4005)$

$\mathtt{Q_{sh} = 600.75 \ vars}$

3 0

3 years ago

Give two examples of human activity that affect the earth systems

joja [24]

Humans affect the eath by burning fossil fuels

4 0

3 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

3 years ago

Why is it harder to get something moving than to keep it moving?

vovikov84 [41]

Answer:

momentum

Explanation:

when something starts rolling momentum keeps it going.

6 0

3 years ago

You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 30.5 ∘ from the ground with a

emmainna [20.7K]

Answer: h = 20.92 m

Explanation: By using the law of conservation of energy, the kinetic energy of the ball equals it potential energy.

Kinetic energy =mv^2/2

Potential energy = mgh

Where m = mass of the object, v = velocity of object = 23.5 m/s

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

mv^2/2 = mgh

m cancels out each other on both sides , hence we have that

v^2 = 2gh.

We want the ball to move towards the wall (horizontal motion), hence we need the horizontal component of the velocity since the velocity is inclined at an angle of 30.5 to the ground (horizontal).

Hence v = 23.5 × cos 30.5, v = 20.248 m/s

Recall that v^2 = 2gh

(20.248)^2 = 2×9.8×h

409.98 = 19.6 h

h = 409.98/ 19.6

h = 20.92 m

7 0

3 years ago