A celestial body moving in an ellipical orbit around a star

Answers

Answer 1

Answer: A celestial body moving in an elliptical orbit around a star is a planet.

Answer 2

Answer:

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

Related Questions

. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.
Please help really easy
A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?
TsAn electric light bulb mixer is used for 19.2 seconds. In that time, it transfers 1536 J of energy.Calculate the power output of the cake mixer.AI DONT KNOW
Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.

A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 ✕ 1023 units.)

Answers

Since volume of each cube is 1 cm^3
Then we can get the volume of 1 mole of cubes, which is $1 * 6.02 * 10^23 cm^3$
The edge $edge = v^1/3$
And the new adge that we are looking for: $new edge = (6.02*10^23)^1/3$ = $= 1.8191 * 46415888.336 = 84435142.472$
So, the final soution for the edge length of cube is 844km.

Do hope it helps!
Regards.

Assume we are given an electric field set up by an unknown charge distribution. U0 is the amount of work needed to bring a point charge of charge q0 in from infinity to a point P. If the charge q0 is returned to infinity, how much work would it take to bring a new charge of 4 q0 from infinity to point P?

Answers

Answer:

$4U_0$

Explanation:

We are given that

Amount of work needed to bring a point charge q0 from infinity to a point P= $U_0$

We know that potential at point P= $V=(U)/(q)$

$U=Vq$

Where U=Amount of work needed to bring a point charge q from infinity to a point P

Initially , $V=(U_0)/(q_0)$

New charge, q= $4q_0$

Then, work done,U= $(U_0)/(q_0)* (4q_0)=4U_0$

Hence, the amount of work needed to bring a new charge 4q0 from infinity to point P= $4U_0$

Answer:

$4U_(0)$

Explanation:

V = u / q,

Work = P = V

1U / 1/4 = 4U

The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.

Answers

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed ,E=Electric filed

Power P = I A

A=Area=4πr² ,I=Intensity

$I=(CB^2)/(2\mu_0)$

$I=(CE^2)/(2\mu_0 C^2)$

$E=\sqrt{{2I\mu_0 C}}$

$E=\sqrt{{2* 1150* 4\pi * 10^(-7)(2.99792* 10^8)}}$

E=930.84 N/C

Therefore answer is 930.84 N/C

Final answer:

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

Explanation:

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

Learn more about Electromagnetic Waves here:

brainly.com/question/29774932

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Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answers

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_(disk) + I _(sand)\n\nI_f = (M_DR^2)/(2) + M_SR^2\n\nI_f = (5*1^2)/(2)+ 3*0.5^2\n\nI_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s

What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

Answers

Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
= (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
= (9.1×10∧-31 × 1,153,846.154∧2) ÷2
= (1.21154×10∧-18) ÷ 2
= 6.05769×10∧-19 J

Answer:

K = 6.02 × 10⁻¹⁹ J

Explanation:

The momentum (p) of an electron is its mass (m) times its speed (v).

p = m × v

v = p / m = (1.05 × 10⁻²⁴ kg.m/s) / 9.11 × 10⁻³¹ kg = 1.15 × 10⁶ m/s

We can find the kinetic energy (K) using the following expression.

K = 1/2 × m × v²

K = 1/2 × 9.11 × 10⁻³¹ kg × (1.15 × 10⁶ m/s)²

K = 6.02 × 10⁻¹⁹ J

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A ship is traveling at 154 m/s and accelerates at a rate of 1.80 m/s^2 for 1 minute. What will its speed be after that minute? Calculate the answer in both meters per second and kilometers per hour.