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3 years ago

For the simple harmonic motion equation

Physics

1 answer:

My name is Ann [436]3 years ago

3 0

Given that : d = 5sin(pi t/4), So, maximum displacement, d = 5*(+1) = 5 Also, maximum displacement, d = 5*(-1) = -5

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SVEN [57.7K]

Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: $f=\frac{v}{\lambda}$

(frequency = velocity/wavelength).

We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

Now we can substitute these values into the formula and calculate to solve:

$f=\frac{340}{0.22} \\\\f=1545.454...$

Simplify to 3 significant figures:

f = 1,550Hz

(Which I believe is just below a G6 if you were interested)

Hope this helped!

4 0

3 years ago

Calculate the minimum thickness (in nm) of an oil slick on water that appears blue when illuminated by white light perpendicular

mr_godi [17]

Answer:

The minimum thickness = 83.92 nm

Explanation:

The relation between the wavelength in a particular medium and refractive index $\lambda_n = \frac{ \lambda }{n}$

where ;

$\lambda$ = wavelength of the light in vacuum

n = refractive index of medium with respect to vacuum

For one phase change :

$2t = \frac{\lambda_n}{2}\\\\where \ \lambda_n = \frac{\lambda}{n}\\\\Then \ \\\\2t = \frac{\lambda}{2n}\\\\t = \frac{\lambda_n}{4n}$

Replacing 1.43 for n and 480 nm for λ; we have:

$t = \frac{480}{4(1.43)}$

t = 83.92 nm

Thus; the minimum thickness = 83.92 nm

4 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

A 4.0-kg object has 72 J of kinetic energy.<br> Determine its speed.

Svetlanka [38]

Answer:

6m/s

Explanation:

v = sqrt of 2KE/m

Where:

KE = kinetic energy

m = mass of a body

v = velocity of a body

= sqrt of 2(72)/4

= sqrt of 144/4

= sqrt of 36

= 6m/s

8 0

2 years ago

A plane starting from rest (vo = 0 m/s) when t0 = 0s. The plane accelerates down the runway, and at 29 seconds, its velocity is

IrinaK [193]

Answer:

Acceleration, $a=2.48\ m/s^2$

Explanation:

Given that,

The plane is at rest initially, u = 0

Final speed of the plane, v = 72.2 m/s

Time, t = 29 s

We need to find the average acceleration for the plane. It can be calculated as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{72.2}{29}$

$a=2.48\ m/s^2$

So, the average acceleration for the plane is $2.48\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago