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The star Betelgeuse is 20 times more massive than the sun. What would be the likely impact on the motion of Earth if the sun wer

Nataliya [291]

Orbital radius would decrease because the gravitational pull would increase.

3 0

4 years ago

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A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.841 rad/s.0.841 rad/s. You, with a mass

kvv77 [185]

Answer:

The total angular momentum of the system is $217.46\ kg-m^2/s$ .

Explanation:

Given that,

Angular speed = 0.841 rad/s

Mass of platform = 72.1 kg

Speed = 1.17 m/s

Mass of poodle = 20.3 kg

Mass of mutt = 18.5 kg

Distance =3/4 of the platform's radius from the center

Mass of disk = 92.5 kg

radius = 1.87 m

Angular momentum is the product of moment of inertia and angular speed.

We need to calculate the angular momentum of person

$L_{person} = I\omega$

$L_{person}=\dfrac{1}{2}mr^2\times\omega$

$L_{person}=\dfrac{1}{2}\times72.1\times(\dfrac{v}{\omega})^2\times\omega$

$L_{person}=\dfrac{1}{2}\times72.1\times(\dfrac{1.17}{0.841})^2\times0.841$

$L_{person}=58.679\ kg m^2/s$

We need to calculate the angular momentum of platform

$L_{platform}=\dfrac{1}{2}mr^2\times\omega$

Put the value into the formula

$L_{platform}=\dfrac{1}{2}\times92.5\times1.87^2\times0.841$

$L_{platform}=136.02\ kg m^2/s$

We need to calculate the angular momentum of poodle

$L_{poodle}=\dfrac{1}{2}mr^2\times\omega$

Put the value into the formula

$L_{poodle}=\dfrac{1}{2}\times20.3\times(\dfrac{1.87}{2})^2\times0.841$

$L_{poodle}=7.4625\ kg m^2/s$

We need to calculate the angular momentum of mutt

$L_{mutt}=\dfrac{1}{2}mr^2\times\omega$

$L_{mutt}=\dfrac{1}{2}\times18.5\times(\dfrac{3}{4}\times1.87)^2\times0.841$

$L_{mutt}=15.302\ kg m^2/s$

We need to calculate the total angular momentum

$L=L_{person}+L_{platform}+L_{poodle}+L_{mutt}$

$L=58.679+136.02+7.4625+15.302$

$L=217.46\ kg-m^2/s$

Hence, The total angular momentum of the system is $217.46\ kg-m^2/s$ .

3 0

3 years ago

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

Karolina [17]

The path difference between two waves is the distance between the two waves. This distance may be obtained as the product of the order of the maxiuma or minima and the wavelength of the wave. The formula is:
d*sin∅ = nλ

Substituting the values for the distance between slits and the angle:
nλ = 0.05 * sin(0.25)
nλ = 2.18 x 10⁻⁴ m

The path difference is 0.2 millimeters.

7 0

3 years ago

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A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is

Snowcat [4.5K]

Answer:

7.8x10-12N

Explanation:

We know that

Magnetic force = F = qVB

And

Also Kinetic energy K.E is

E = (1/2)mV²

So making v subject

V = √(2E / m)

And

E = KE = 2MeV

= 2 × 106 eV

= 2 × 106 × 1.6 × 10–19 J

= 3.2 × 10–13 J

And then

V= √2x3.2E-13/1.6E-27

1.9E7m/s

Given that

mass of proton = 1.6 × 10–27 kg,

Magnetic field strength B = 2.5 T.

So F= qBv sinစ

=

So F = 1.6 × 10–19 × 2.5 × 1.9 x10^7 x sin 90°

= 7.8 x 10^-12N

5 0

3 years ago

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A magnetic field applies forces on: a)static charges b)moving charges c)water flow

lisabon 2012 [21]

Answer:

moving charges b)

I think this is the answer

4 0

4 years ago

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