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2 years ago

5

Which new process was added to the nebular theory of solar system formation to account for the existence of Jupiter-mass planets

that orbit close to their stars (the "hot Jupiters")?

1 answer:

Nikolay [14]2 years ago

7 0

Answer:

Explanation:

Due to their very intense activity and magnetic fields, they are covered with sunspots hundreds of times bigger than those on the Sun. These therefore generate perturbations in the star's spectrum that are much greater than those caused by planets, making them much harder to detect, even when they are hot Jupiters.

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What type of light does this light bulb produce most (i.e. at what wavelength does the spectrum have maximum intensity)?

lesantik [10]

Answer: The light bulb produces the continuous light. At minimum wavelength the spectrum have maximum intensity.

Explanation:

According to Wein's displacement law, the wavelength is inversely proportional to the temperature.

The intensity depends on the frequency. The frequency is inversely proportional to the wavelength.

Therefore, when the temperature of the light bulb will be maximum then the wavelength will be minimum. At minimum wavelength the spectrum have maximum intensity.

4 0

2 years ago

Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

Nata [24]

The frequency of a sound is whatever frequency leaves the source. It doesn't change.

Voiced of swimmers at the pool don't change frequency in or out of the water. Only their speed and wavelength change.

5 0

2 years ago

Read 2 more answers

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

arsen [322]

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

V = 0.2977 m/s

The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.

7 0

2 years ago

A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = $\frac{\pi d^{2} }{4}$

where d is the diameter of the wire

1.613 x 10^-7 = $\frac{3.142* d^{2} }{4}$

6.448 x 10^-7 = 3.142 x $d^{2}$

$d^{2}$ = $\sqrt{ 2.05*10^-7}$

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0

2 years ago