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3 years ago

Which of the following is true about a planet orbiting a star in uniform circular

1 answer:

7 0

The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.

A. The velocity vector of the planet points toward the center of the circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.

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Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

2 years ago

When Dr. Hewitt immerses an object in water the second time and catches the water that is displaced by the object, how does the

Dimas [21]

Answer:

Explanation:

- The volume of water displaced by immersing the object is equal the amount of water spilled and caught by Dr. Hewitt.

- The amount of water is proportional to the volume of object of fraction of object immersed in water will lead to the same fraction of water displaced and caught by Dr. Hewitt.

- When the object is immersed the force of Buoyancy acts against the weight and reducing the scale weight.

- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.

5 0

2 years ago

True or False: <br><br> Solids always<br> have a higher density than<br> liquids and gases.

Hunter-Best [27]

Answer:

<h3>False</h3>

<h3> I hoped this helpful for you....</h3>

Thank you ☺️☺️

7 0

3 years ago

Read 2 more answers

A ring placed along y 2 + z 2 = 4, x = 0 carries a uniform charge of 5 μ C/m. ( a ) Find D at P(3,0,0) . ( b ) If two identical

GuDViN [60]

For right now idk :p

3 0

2 years ago

A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's

Dmitrij [34]

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0

2 years ago