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1 year ago

Suppose an asteroid had an orbit with a semimajor axis of 4 au. how long would it take for it to orbit once around the sun?

Physics

1 answer:

Nitella [24]1 year ago

5 0

It would take <u> 8 years </u> for the asteroid to orbit once around the sun.

What is a semimajor axis?

In geometry, the major axis of an ellipse is its longest diameter: a line segment that runs through the center and both foci, with ends at the two most widely separated points of the perimeter.
The semi-major axis (major semiaxis) is the longest semidiameter or one half of the major axis, and thus runs from the centre, through a focus, and to the perimeter.
The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches.
Thus it is the distance from the center to either vertex of the hyperbola.
In astronomy, the semi-major axis is one of the most important orbital elements of an orbit, along with its orbital period.
For Solar System objects, the semi-major axis is related to the period of the orbit by Kepler's third law.

To know more about semi-major axis, refer:

brainly.com/question/26662489

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Calculate the force exerted on this test dummy with a mass of 75 kg hits an air bag accelerating at 12 m/s2.

maksim [4K]

Answer: 900

Explanation: Force equals mass x acceleration F=M×A

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3 years ago

PLEASE HELP!!!! The displacement vectors A and B, when added together, give the resultant vector R, so that R = A + B. Use the d

GuDViN [60]

Answer:

Ax = 0

Ay = 6 m

Bx = 8 cos phi = cos 34 = 6.63 m

By = 8 sin phi = 8 sin (-34) = -4.47 m

Rx = Ax + Bx = 0 + 6.63 = 6.63 m

Ry = Ay + By = 6 - 4.47 = 1.53 m

R = (6.63^2 + 1.53^2)^1/2 = 6.80 m

tan theta = Ry / Rx = 1.53 / 6.8 = ,225

theta = 12.7 deg

7 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

A circular loop of wire is in the plane of the paper. The south pole of a bar magnet is being moved from a position in front of

drek231 [11]

Answer:

Counterclockwise

explanation in attachment

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3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

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