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kakasveta [241]

3 years ago

5

Which statement is true about acceleration?

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

The answer is C is think

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A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9

Black_prince [1.1K]

Answer:

$C = 46.891 \mu F$

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

$s = \frac{P}{P.F} < COS^{-1} 0.65$

$S = \frac{1250}{0.65} < 49.45$

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

$P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]$

solving for $Q_{new}$

$Q_{new} = P tan [cos^{-1} P.F new]$

$Q_{new} = 1250 [tan[cos^{-1}0.9]]$

$Q_{new} = 605.40 VARS$

$Q_C = Q - Q_{new}$

$Q_C = 1461 - 605.4 = 855.6 vars$

$Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C$

$C = \frac{Q_C}{ v_{rms}^2 \omega}$

$C = \frac{855.6}{220^2 \times 2\pi \times 60}$

$C = 4..689 \times 10^{-5}$ Faraday

$C = 46.891 \mu F$

6 0

4 years ago

What are the methods of storing tools

irina [24]

You might want to visit this website
https://www.nedcc.org/free-resources/preservation-leaflets/4.-storage-and-handling/4.1-storage-metho...

8 0

3 years ago

A rocket is fired at 100 m/s at an angle of 37, how many seconds is it in the air?

Sonbull [250]

Answer:

12.3 seconds

Explanation:

time in air = 2(v)/g

2(100sin(37))/9.8

=12.28 seconds

7 0

4 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

F = $m \times a$

= $1090 \times (\frac{21}{16})$

= 1430.625 N

Also, it is known that

$sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}$

$sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}$

$\theta$ = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

8 0

3 years ago

A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must

Sergio039 [100]

Explanation:

The centripetal force $F_c$ on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

$y:\:\:\:\:N - mg = 0$

$x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N$

or

$m \dfrac{v^2}{r} = \mu mg$

Solving for $\mu$ ,

$\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28$

3 0

3 years ago