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satela [25.4K]
4 years ago
10

Velocity is a vector quantity which has both magnitude and direction. ... Net force is also a vector quantity which has both mag

nitude and direction. Using complete sentences, describe the net force acting upon the object during the course of the elliptical orbit. Comment on both the magnitude and the direction.
Physics
2 answers:
GuDViN [60]4 years ago
7 0
When the object is moving in the elliptical orbit, it means that the direction of its acceleration should be towards the two foci (plural of focus) of the ellipse to keep the elliptical motion. As force according to the Newton's second law: F = ma, the net force must be in the direction of the acceleration. As far as the magnitude of net force is concerned, you can use Newton's gravitational law to find its magnitude.
Lera25 [3.4K]4 years ago
3 0

The net force on an object moving in an elliptical path is the gravitational force. The force vector always points towards the object being orbited by the object following the elliptical orbit. In the case of the sun and earth, the force vector will always point towards the sun

The exact magnitude of the force is given by Newton's law of universal gravitation. The formula to calculate the magnitude of this force is ,F=\frac{GMm}{r^2}.

In this formula G is the universal gravitational constant, r is the distance between the two objects, M,m each represent the masses of the central object and the object orbiting it.  As a consequence we can tell that closer to the central object, the magnitude of the force will be larger than further away from the object.

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Please help, I don’t get the question.
aalyn [17]

All you have to do is get the culmative effect of the forces by predicting the direction in which each object will move. Just write down what you know and put down questions for your teacher.

6 0
3 years ago
Lab: Motion with Constant Acceleration Assignment: Lab Report
matrenka [14]

Answer

3

Explanation:

x=v_{0}<em>t + 1/2 </em>at^{2}<em />

3 0
3 years ago
A force of 30 N is exerted on an object on a frictionless surface for a distance of 6.0 meters. If the object has a mass of 10 k
tangare [24]

Answer:The velocity of the object will be 5\sqrt7m/s or 13.23m/s

Explanation:

force exerted by the object= 30N

distance displayed by the object by the action of force=6.0m

mass of object=10kg

velocity gained by the object=?

\frac{1}{2}mv^{2}= forcexdisplacement\\\frac{1}{2}10v^{2} = 30x6\\ 5v^{2}=180\\ v^{2}= 180-5\\ v^{2} =175\\v=\sqrt{175} \\v=5\sqrt{7} or 13.23

6 0
2 years ago
Please answer this question given in the picture
elena-s [515]

A plane mirror forms a virtual image behind the mirror. The image is as far behind the mirror as the object is in front of it. A cannot see his image because the length of the mirror is too short on his side. However, he can see the objects placed at points P and Q, but cannot see the object placed at point R



Hope this helps Buddy!


~ Courtney

6 0
4 years ago
A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 205 Hz. A person on the pl
Arlecino [84]

Answer:

Explanation:

The question relates to Doppler effect and beat.

The observer is moving towards the reflected sound so apparent frequency will be increased

f = f₀ x (V + v₁) / (V - v₂)

f is apparent frequency , f₀ is real frequency , V is velocity of sound , v₁ is velocity of observer and v₂ is velocity of source . Here

v₁ = v₂ = vp as both observer and source have same velocity

f = f₀ x (V + v₁) / (V - v₂)

205 +5 = 205 x (344 +vp)/ ( 344 - vp)

1.0234 = (344 +vp)/ ( 344 - vp)

= 352 - 1.0234vp = 340+vp

12 / 2.0234vp

vp = 6 m /s approx.

6 0
3 years ago
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