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vlabodo [156]
3 years ago
14

satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is close

st to the Earth, it is a distance of1.00×106m from the surface (not the center) of the Earth, andis moving at a velocity of5.14km/s. At the point in its orbit when it is furthest from the Earth it is a distance of2.00×106m from the surface of the Earth. Note that the Earth has a mass of5.97×1024kgand a radius of6.37×106m.(a)How fast is the satellite moving when it is at its furthest point from the Earth?(b)If, at the closest approach to the Earth, the satellite could be deflected so that it had the same velocity as before,but was now traveling directly away from the Earth, how far from the surface of the Earth would it get before itstopped, and then started to fall back to the Earth?
Physics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m

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Five difference between elastic collision and inelastic collision?​
olga_2 [115]

Answer:

Elastic Collision

Inelastic Collision

The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.

Momentum does not change. Momentum changes.

No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.

Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.

An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.

8 0
3 years ago
In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es
tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0
3 years ago
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
7 0
3 years ago
When work is done, an object changes its
STatiana [176]

Answer:

energy is added to it

Explanation:

8 0
3 years ago
2. Light waves of the wavelength of 650 nm and 500 nm produce interference fringes on a screen at a distance of 1m from a double
satela [25.4K]

Answer:

least distance= 13mm

ratio of the lattice = 1 : 0.71 : 0.58

Explanation:

given λ₁ = 650nm = 650×10⁻⁹m,  λ₂ = 500nm = 500×10⁻⁹m

5 0
3 years ago
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