3 years ago

Which component of the galaxy is shown in this image? Dust

Physics

1 answer:

Westkost [7]3 years ago

3 0

Answer:

stars

those are stars in the galaxy

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A person of mass 80 kg stands at the center of a rotating merry-go-round platform of radius 3.5 m and moment of inertia 950 kg*m

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Answer with Explanation:

We are given that

Mass of person,M=80 kg

Radius,r=3.5 m

Moment of inertia of merry,I= $950 kgm^2$

Angular velocity of platform= $\omega=0.85rad/s$

1.Law of conservation of angular momentum

$I\omega=I'\omega'$

Where $I'=950+80(3.5)^2$

Substitute the values

$950\times 0.85=(950+80(3.5)^2)\omega'$

$\omega'=\frac{950\times 0.85}{950+80(3.5)^2}=0.418rad/s$

Angular velocity when the person reaches the edge =0.418 rad/s

2.Rotational kinetic energy of the system before the person's walk

$K_i=\frac{1}{2}I\omega^2=\frac{1}{2}(950)(0.85)^2=343.2 J$

3.Rotational kinetic energy of the system after the person's walk

$K_f=\frac{1}{2}(950+80(3.5)^2)\times (0.418)^2$

$K_f=168.6 J$

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3 years ago

In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- ve

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Answer:

$B_{int}=-0.015T$

Explanation:

From the question we are told that:

RF source oscillation speed $\sigma= 34 MHz$

The external field $Bext =0.78 T$ .

Pro- ton magnetic moment component $\mu=1.41 X 10-26 J/T$

Generally the equation for magnitude of $B_{int}$ is mathematically given by

$B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}$

$B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}$

$B_{int}=0.78-0.7957$

$B_{int}=-0.015T$

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Which is a characteristic of all ions? They are made of one type of atom. They have one overall charge. They are made of two or

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Answer:

They are made of one type of atom.

Explanation:

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A particle of charge of +3.40 ✕ 10−6 C is 17.5 cm distant from a second particle of charge of −2.00 ✕ 10−6 C. Calculate the magn

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Answer:

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Explanation:

q1 = 3.4 x 10^-6 C

q2 = 2 x 10^-6 C

d = 17.5 cm = 0.175 m

The electrostatic force is given by

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F = 0.47 N

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Decrease the amount of work done.

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