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2 years ago

Also known as the dog star, what is special about sirius?

1 answer:

6 0

Sirius, also known as the Dog Star or Sirius A, is the brightest star in Earth's night sky. The name means "glowing" in Greek — a fitting description, as only a few planets, the full moon and the International Space Station outshine this star. Because Sirius is so bright, it was well-known to the ancients.

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A 0.350kg bead slides on a curved fritionless wire,

LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

$m1*g*h-\frac{m1*V_B^2}{2}=0$ Solving for Vb:

$V_B=\sqrt{2gh}=6.56658m/s$

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

$V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s$

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

$m1*g*h2-\frac{m1*V_{B'}^2}{2}=0$ Solving for h2:

h2 = 0.092m

6 0

3 years ago

Use the diagram to explain how convection occurs inside the earth. What could convection cause to occur on earths surface

liubo4ka [24]

Answer:

mantle convection is the very slow creeping motion of earths solid silicate mantle caused by convection currents carrying heat from the interior to the planet's surface.

3 0

3 years ago

A cat with a mass of 4.50 kilograms sits on a ledge 0.800 meters above the ground. If it jumps to the ground, how much kinetic e

Slav-nsk [51]

Answer:

The cat will have 36J of kinetic energy.

5 0

3 years ago

Read 2 more answers

A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

Kazeer [188]

Answer:

$\Delta K = 52J$

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: $\Delta K=K_f-K_i$

We know that the formula for the kinetic energy for an object is:

$K=\frac{mv^2}{2}$

where m is the mass of the object and v its velocity.

For our case then we have:

$\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}$

Which for our values is:

$\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J$

3 0

3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago