What distance is moved if we have a 8N force and the work done is 90J

Suppose you found a mineral crystal that looked like topaz. what two minerals could you use in a scratch test to help determine

Vesna [10]

The two most important minerals you need to do the test are <span>Corundum and quartz. You use that to do scratch test</span>

4 0

3 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

Why is it advisable to wear long sleeves when a student works in a chemistry lab

love history [14]

So then if they do spill the chemical then it gets on their cloths and not on then it dosen"t harm them instead it ruins their shirt.

3 0

3 years ago

Read 2 more answers

Two trains are headed towards each other on the same track unbeknownst to the engineers. One departs San Francisco. Its average

aalyn [17]

Answer:

7,166 hrs =430 minutes

Explanation:

Since both train are on the same track, going one towards the other, the relative speed is the addition of both, then the time they need to meet, and consistently crash, is the time that (65mph + 55 mph)=120mph need to travel the total distance of 860 miles, of course in this case one part is traveled by the first train and the rest by the other. Then to find the time we use a three rule

1 h --->120mi

X ---->860mi, then X=(860 mi* 1h)/120 mi = 43/6 hrs= 7,16666 hrs, turning this into minutes need that we notice 1h=60min, then 43/6 hrs *60 min/hrs = 430 minutes.

7 0

3 years ago

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

alexdok [17]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

$X=V_{o}*t + \frac{a*t^{2}}{2}$ Replacing the known values:

$1500=6t+0.9*t^{2}$ Solving for t we get 2 possible answers:

t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

8 0

3 years ago