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2 years ago

What is the relationship between mass and weight?

Physics

1 answer:

Vikentia [17]2 years ago

4 0

Answer:

C. Weight is mass times the acceleration of gravity.

Explanation:

Weight is the force exerted by gravity. Force is equal to mass times acceleration.

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You replace a 100 W incandescent lightbulb with a 22 W CFL bulb. If you leave your lights on 24 hours a day, how much energy are

crimeas [40]

100 watts = 100 joules per second
22 watts = 22 joules per second
Saving = 78 joules per second

1 hour = 3,600 seconds
24 hours = 86,400 seconds

(78 joules/sec) x (86,400 sec) = 6,739,200 joules

5 0

2 years ago

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!

Igoryamba

C. FM radio waves is the answer.

6 0

2 years ago

I need help like right now!!!!! please

Mariana [72]

1. All of the above

2. Lack of consistent sunlight

3. Nonpolluting & Can be developed anywhere

4. Protons and Neutrons

sorry the answer is so late...Hope it still helps u

Also I'm pretty sure all the answers I provided are correct but I'm not sure 4 a fact so plz let me know...

5 0

3 years ago

The successive ionization energies for an unknown element are listed below. To which family in the periodic table does the unkno

bagirrra123 [75]

The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.

So there will be only 2 electrons in the outer most shell.

According to the information mentioned above we can conclude the unknown element likely belongs to the second group.
I2 = 1752 kj/mol

4 0

2 years ago

Read 2 more answers