+ CaCl 2 + _C2HCl3 C2H2C14 H 20 + Ca(OH) 2 →

Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the

mixas84 [53]

Answer:

$T=1.54s$

Explanation:

From the question we are told that:

Speed of Motor 1 $\omega_1=825rpm=>2 \pi 13.75$

Speed of Motor 2 $\omega_1=786rpm=>2 \pi 13.1$

Therefore

Frequency of Motor 1 $f_1=13.75$

Frequency of Motor 2 $f_2= 13.1$

Generally the equation for Time Elapsed is mathematically given by

$T=\frac{1}{df}$

Where

$df=f_1-f_2$

$df=13.75-13.1$

$df=0.65Hz$

Therefore

$T=\frac{1}{65}$

$T=1.54s$

5 0

3 years ago

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s

yarga [219]

Answer:

$I=336.6kgm/s$

Explanation:

The equation for the linear impulse is as follows:

$I=F\Delta t$

where $I$ is impulse, $F$ is the force, and $\Delta t$ is the change in time.

The force, according to Newton's second law:

$F=ma$

and since $a=\frac{v_{f}-v_{i}}{\Delta t}$

the force will be:

$F=m(\frac{v_{f}-v_{i}}{\Delta t})$

replacing in the equation for impulse:

$I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)$

we see that $\Delta t$ is canceled, so

$I=m(v_{f}-v_{i})$

And according to the problem $v_{i}=0m/s$ , $v_{f}=5.10m/s$ and the mass of the passenger is $m=66kg$ . Thus:

$I=(66kg)(5.10m/s-0m/s)$

$I=(66kg)(5.10m/s)$

$I=336.6kgm/s$

the magnitude of the linear impulse experienced the passenger is $336.6kgm/s$

6 0

3 years ago

Please help with this question! Thanks in advance...

-Dominant- [34]

A.0.6 hope you do good on your test!!

5 0

3 years ago

A rock is dropped from a tall bridge and hits the water below in 2.9

Ahat [919]

Answer:

The height of the bridge is 14.5

Explanation:

Given the following data

t = 2.9 seconds

g = +10m/s^2

Using the below formula

H = ut + 1/2gt^2

Since the initial velocity u = 0

Then, H = 1/2gt^2

H = 1/2 x 2.9 x 10

H = 2.9 x 10 / 2

H = 2.9 x 5

H = 14.5

3 0

3 years ago

An AC circuit has an RMS voltage of 80 VAC. What's the circuit's average voltage?

yaroslaw [1]

-- The RMS value of an AC waveform is (1/2)(√2) x (peak value)

So the peak value is (√2) x (RMS value)

-- The Average value of an AC waveform is (2/π) x (peak value)

So the peak value is (π/2) x (Average value).

-- So far, this is all very entertaining, but how does it help us answer the question.

Well, we found the peak value in terms of the RMS and in terms of the Average. So we can set these equal to each other, and solve for the Average in terms of the RMS. This sounds like such a good plan, I think I'll do it !

Peak = (√2) x (RMS value) and Peak also = (π/2) x (Average value).

So (√2) · RMS = (π/2) · Average .

Divide each side by π : (√2) · RMS / π = (1/2) · Average

Multiply each side by 2 : Average = (2/π) · (√2) · RMS .

You said that the RMS value is 80 V, so

Average = (2/π) · (√2) · (80)

Average = (2 · √2 / π) · (80)

Average = 160√2 / π

Average = 72 volts . (But be sure to read the 'gotcha' below.)

Now I'll go ahead and tell you the 'gotcha':

All of these numbers are true, as far as they go. But the 'average' is only true for 1/2 cycle of an AC wave. Picture an AC wave in your mind. You'll see that it spends just as much time being negative as it spends being positive. So the 'average' of any number of AC whole cycles is zero.

7 0

3 years ago