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2 years ago

What is the amplitude of an AC voltage waveform, in units of Volts, if the RMS value is 369 V?

Physics

1 answer:

Blizzard [7]2 years ago

4 0

Answer:

521.8 V

Explanation:

The RMS value of the voltage of an AC signal is given by

$V_{rms}=\frac{V_0}{\sqrt{2}}$

where

$V_0$ is the peak voltage, which corresponds to the amplitude of the AC waveform

In this problem, we know the RMS voltage

$V_{rms}=369 V$

Therefore, we can re-arrange the previous equation to find the peak voltage (the amplitude of the waveform):

$V_0 = \sqrt{2}V_{rms}=\sqrt{2}(369 V)=521.8 V$

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Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

2 years ago

Please help easy 8th grade question

Tamiku [17]

Answer:

whats the question?

Explanation:

7 0

3 years ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago

An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a

Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile $a=3m/sec^2$

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So $v=5.13+3\times 4.9=19.83m/sec$

So the final velocity is 19.83 m/sec

7 0

2 years ago

Identify which best describes the energy used to pluck guitar strings to make sound.

Novay_Z [31]

<span>The correct answer is: Mechanical Energy

Explanation:
As the guitar strings are plunked, the potential energy stored in the strings has an ability to make them vibrate. When the strings are vibrating, that potential energy is actually converted to the kinetic energy. Hence, the whole phenomena contains both the kinetic energy and the potential energy. The sum of kinetic energy and the potential energy is called Mechanical energy. Therefore, the correct answer is Mechanical Energy.</span>

8 0

3 years ago

Read 2 more answers