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aliina [53]
3 years ago
5

I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha

t was my reaction time)?
Physics
1 answer:
statuscvo [17]3 years ago
3 0
D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

acceleration=acceleration due to gravity=-9.8m/s^2

It falls - 22cm or -0.22m

We have - 0.22=1/2(-9.8)t^2

t^2=(-0.44)/(-9.8)

t=sqrt[0.44/9.8]
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Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

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In Young’s two slit experiment, the first dark fringe above the central bright fringe occurs at an angle of 0.44˚. What is the r
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Answer:

d / λ = 26.7

Explanation:

In Young's double slit experiment, constructive interference is described by the expression

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In the case of destructive interference we must add half wavelength (λ/2)

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