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2 years ago

Can someone write and short summary on how to hula hoop

2 answers:

5 0

You pick up the hula hoop and stand inside of it then pick it up will you’re inside it and hold to your waist and spin it then turn your hips

MissTica2 years ago

5 0

Answer:

First you take a hula hoop and hold it around your waist then you start moving your hips in a circle and let the hula hoop go off your hands and you keep moving your hips. You keep moving your hips until you want to stop but it might fall. Just keep trying and you should get the hang of it in a few minutes.

Explanation:

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Two transverse waves travel along the same taut string inopposite directions. the waves are described by following equations use

Umnica [9.8K]

Answer: y'=2Asin(kx)cos(wt)

Explanation:

Let y1=A sin (kx + wt) be the first wave

y2=A sin (kx - wt) be the second wave in the opposite direction (which we showed by putting a negative sign between the terms kx and wt)

Please do note that both wave have the same attributes (that's Amplitude, wave number and angular frequency) because they are formed on the same medium by the same source just that their directions are opposite.

By super imposing these 2 waves, we have a resulting singular wave representing both wave (law of superimposition) with a resulting value of vertical displacement y'.

Thus y' = y1 + y2.

Let us do the math.

y'=A sin (kx + wt) + A sin (kx - wt)

By factoring A out, we have that

y' = A [ sin (kx + wt) + sin (kx - wt)]

For simplicity let us use the substitution

Let (kx + wt) = a and (kx - wt) =b

Hence we have that

y' = A [sin a + sin b].

From trigonometric ratio

sin a + sin b = 2sin[(a+b)/2] * cos [(a - b)/2]

By recalling that (kx + wt) = a and (kx - wt) =b

sin a + sin b = 2sin [(kx +wt +kx-wt) /2] * cos [(kx +wt - (kx-wt))/2]

Thus we have that

sin a + sin b = 2sin [(kx+wt+kx-wt)/2] * cos[(kx+wt-kx+wt)/2]

By collecting like terms in the bracket we have that

sin a + sin b = 2sin[2kx/2] * cos [2wt/2]

By dividing

sin a + sin b = 2sin(kx) cos(wt)

Now let us get the final resultant vertical displacement (y')

Recall that

y' = A [sin a + sin b]. and we already deduced that

sin a + sin b = 2sin(kx) cos(wt)

Finally,

y' = A [2sin(kx) cos(wt)] which is

y'=2Asin(kx)cos(wt)...... Final answer

4 0

3 years ago

As the frequency of a wave increases the wavelength.

zaharov [31]

Answer:

decreases

Explanation:

Remeber:

There is always inverse relation between frequency and wavelength.

So if one of them increases, other decreases and vice-versa.

f ∝ 1 / λ

4 0

1 year ago

Read 2 more answers

How would you write the number 6,500,000,000 in scientific notation?

photoshop1234 [79]

Scientific form = 6.5 x 109.

8 0

3 years ago

Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p

Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation that is

for red light

h ( c / λr ) = Ф + kinetic energy

for green light

h ( c / λg ) = Ф + 1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)

put all value

Ф =( 6.63 × $10^{-34}$ (3 × $10^{8}$ ) / 0.50) × ( 1.50/ λr - 1/ λg)

Ф =( 6.63 × $10^{-34}$ (3 × $10^{8}$ ) / 0.50 ) × ( 1.50/ 651× $10^{-9}$ - 1/ 521 × $10^{-9}$ )

Ф = 1.5305 × $10^{-19}$ J × ( 1ev / 1.6 × $10^{-19}$ J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0

2 years ago

Read 2 more answers

When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,

Andre45 [30]

Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

$P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A$

(b) Again the power is given by :

$P=\dfrac{V^2}{R}$

R is resistance of the dryer

$R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega$

(c) The magnetic field produced by the dryer at the user's hand is given by :

$B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T$

Hence, this is the required solution.

4 0

2 years ago