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Vikki [24]
3 years ago
9

Starting from rest, a crate of mass m is pushed up a frictionless slope of angle theta by a horizontal force of magnitude F. Use

work and energy to find an expression for the crate's speed v when it is at height h above the bottom of the slope. Express your answer in terms of the variables m, F, theta, h, and free fall acceleration g v=
Physics
1 answer:
ludmilkaskok [199]3 years ago
6 0

Answer:

v =  sqrt[2*(F*h*cot(theta)-mgh)/m]

Explanation:

Work  = KE + Ug

F*r=1/2mv^2+mgh

1/2mv^2=F*r-mgh

v=sqrt[2(F*r-mgh)/m]

r=h/tan(theta)=h*cot(theta)

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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
When light falls on objects, it interacts with them in different ways as shown in the images. What type of interaction is seen i
agasfer [191]

Answers with explanation

1)

Transmission

If all the light passes through a medium without any absorption then transmittance is 100%.

2)

Refraction

Refraction is the bending of a wave when it enters a medium where its speed is different and rays are again refracted when they leave that medium,

3)

Reflection

Reflected rays don't pass through the medium instead rays bounces off an object at an angle.

7 0
3 years ago
What is the tensile strength of hair ?
Olegator [25]
Human hair

Yield strength (MPa)
140-160

Ultimate tensile strength (MPa)
200-250

Hope that helps.
6 0
3 years ago
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A spindle connects two wheels with fixed axles by a fan belt. If the effort wheel is larger than the resistance wheel, which axl
MariettaO [177]
Power = Iω (constant) as they are connected together, since effort axle has large radius than resistance axle, so moment of inertia of effort axle is also more as compared to resistance axle, so angular speed of effort axle is less than the resistance axle. So answer is B. resistance axle  will have more angular speed as its moment of inertia is less for the same power.
8 0
3 years ago
The CIA investigated hypnosis as a possible tool for interrogating prisoners. Why did they decide it was unsuitable for that
alukav5142 [94]

Answer:

D. Hypnosis can make the subjects talk, but they talk only about their childhoods.

Explanation:

7 0
3 years ago
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