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Vsevolod [243]
3 years ago
5

A progressive wave is represented by the following equation :

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Explanation:

I. Direction - Towards the negative x axis since there's a '+' sign b/n π/4 & X

II. y = Asin(wt + kx)

k = 2π/ለ = 1

ለ = 2πcm = 6.29cm or 6.29×10^-2m in meter

i suggest using the one in cm.

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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

Explanation:

Given that,

Diameter = 10 light year

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Using formula of radius

r=\dfrac{d}{2}

r=\dfrac{15\times9.461\times10^{15}}{2}

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M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}

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5 0
3 years ago
Could an experiment similar to young's two-slit experiment be performed with sound? how might this be carried out? does it matte
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7 0
3 years ago
A car moving has a speed of 35 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?
Pepsi [2]

Answer:

7

Explanation:

7 0
3 years ago
Read 2 more answers
A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

   The potential energy possessed by the  hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

        The kinetic energy can be mathematically represented as

                    KE = \frac{mv^2}{2} + \frac{Iw}{2}

Where I is the moment of inertia possessed by the hoop  which is mathematically represented as

                 I = mr^2

Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

                          w = \frac{v}{r}

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

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               mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}

=>            mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}  

=>          mgh = \frac{mv^2}{2} + \frac{mv^2}{2}

=>           mgh = mv^2

=>              v = \sqrt{gh}

Substituting values

                v = \sqrt{9.81 * 1.7}

                  v=4.08m/s

4 0
3 years ago
Read 2 more answers
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