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2 years ago

The minimum amount of energy that has to be added to start a reaction is the ______________ energy.

1 answer:

7 0

The best and most correct answer among the choices provided by your question is the first choice or letter A.

<span>The minimum amount of energy that has to be added to start a reaction is the activation energy.</span>

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1. At a location in Europe, it is necessary to supply 1000 kW of 60-Hz power. Only power sources available operate at 50 Hz. It

Klio2033 [76]

Answer:

Explanation:

From the given information:

The speed of a synchronous motor in relation to its frequency can be represented with the formula:

$n_{sm}= \dfrac{120f_{se}}{P}$

where,

the electrical frequency $f_{se}$ is measured in Hz

the number of poles = P

For us to estimate the number of poles to have 50 Hz - 60 Hz Power, then we need to relate the frequencies of the above equation.

i.e

$\dfrac{120(50 \ Hz)}{P_1}= \dfrac{120( 60 \Hz)}{P_2} \\ \\ \dfrac{6000 \ Hz}{P_1}= \dfrac{7200 \ Hz}{P_2} \\ \\ \dfrac{P_2}{P_1}=\dfrac{7200}{6000} \\ \\ \\ \dfrac{P_2}{P_1}= \dfrac{12}{10}$

Thus, we can conclude that 10 poles synchronous motor is attached with 12 poles synchronous generator in order to convert 50 Hz to 60 Hz power.

3 0

3 years ago

Convert 5g/cm^3 into kg/m^3

Sliva [168]

Answer:

0.01135624

Explanation:

7 0

2 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

2 years ago

Which statement best describes how the period and frequency of electromagnetic waves change between gamma rays and microwaves?

Dimas [21]

On an electromagnetic spectrum, one of its noticeable trends is that the wavelength increases with decreasing energy and the wavelength decrease with increasing energy. Furthermore, gamma rays have high energy and short wavelengths while microwaves have low energy and long wavelengths.

6 0

3 years ago

Read 2 more answers

. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to

AveGali [126]

No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels

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2 years ago