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3 years ago

11

Is a fan a complex machine

1 answer:

shutvik [7]3 years ago

4 0

Hi

Yes, a fan is a complex machine.

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An object has a weight of 500 on earth what is the mass of this object

erastova [34]

550! OBVY! lol! ope this helps1

7 0

3 years ago

On the fashion trends of the 80s where would you begin to look for information

Marrrta [24]

Usually start on the internet, there is bound to be something or a form of information on it.

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3 years ago

A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

6 0

3 years ago

A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio

Step2247 [10]

The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

$W=p\Delta V = p(V_f - V_i)$

where

p is the pressure of the gas

$V_i$ is the initial volume

$V_f$ is the final volume

For the gas in the cylinder in this problem,

p = 2.00 atm

$V_i = 0.250 L$

And we also know the work done,

W = 288 J

So we can solve the equation for $V_f$ , the final volume:

$V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L$

Learn more about ideal gases:

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#LearnwithBrainly

7 0

3 years ago

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

7 0

3 years ago