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poizon [28]
3 years ago
5

The earth’s orbit is oval in shape . Explain how the magnitude of gravitational force between earth and sun changes as earth mov

es from position P to Q as shown in the figure.

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer:

The gravitational force does change believe it or not, but the explaination for this is because the earths orbit is an oval (or a not circle) the closer it nears its self to the sun. the suns gravitational pull, pulls the earth to it bringing it closer and as it reaches the other side lets call this L when it reaches L it becomes the same gravitational pull reset as P is and another section over the sun lets call this M and when it reaches M its the same pull as Q, you get it now?

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Mazyrski [523]
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Hope this helped
6 0
3 years ago
A small bead with a positive charge q is free to slide on a horizontal wire of length 4.5 cm . At the left end of the wire is a
-Dominant- [34]

Answer:

1.5cm

Explanation:

The place where the bead will come to rest is the place where the force between the left charge and the bead is the same, but in opposite direction, as he force between the right charge and the bead.

F_{left} = F_{right}\\K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}

Also:

r_{left}+r_{right}=0.045m\\r_{right} = 0.045m-r_{left}

q_{left}= q\\q_{right}=4q

So, we replace and simplify. Notice that the charges and the Coulomb constant will cancel:

K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}\\\frac{q*q}{r_{left}^2} = \frac{4q*q}{(0.045m-r_{left})^2}\\(0.045m-r_{left})^2 =4r_{left}^2\\0.002025m - 2*0.045r_{left} + r_{left}^2 = 4r_{left}^2\\3r_{left}^2 + 0.09r_{left} - 0.002025m = 0

r_{left} = \frac{-(0.09)+-\sqrt{(0.09)^2-4*(3)(-0.002025)}}{2(3)}\\ r_{left} = 0.015m| -0.045m

But the only solution that would place the bead on the wire is 0.015m or 1.5cm, so this is our answer.

5 0
3 years ago
A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 550 cubic centimeters of soup. The
myrzilka [38]

A microwaveable cup-of-soup package needs to be constructed in the shape of a cylinder to hold 600 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.02 cents per square centimeter. The top will be made of glued paper, costing 0.05 cents per square centimeter. Find the dimensions for the package that will minimize production costs.

h: height of the cylinder, r: radius of the cylinder

The volume of a cylinder: V=πr2h

Area of the sides: A=2πrh

Area of the top/bottom: A=πr2

The cost of packaging, C=2πrh*0.02+ πr^2*0.02+ πr^2*0.05 subject to the constraint πr^2h=600

C=πr(0.04h+.07r)  and the constraint implies h=600/ πr^2

So C=πr(24/πr^2+.07r)=24/r+.07πr^2

C'=-24/r^2+0.14πr=0

r^3=24/0.14π  r=3.79 cm

h=600/πr^2=13.3 cm

C=π*3.79*(0.04*13.3+.07*3.79)=9.48cents

C''=0.14π+48/r^3>0 for all r>=0 so our solution is indeed a minimum.

Learn more about radius at

brainly.com/question/24375372

#SPJ4

7 0
2 years ago
Which particle rarely interacts with matter?
Marysya12 [62]
D. Neutrino
Neutrinos are particles that rarely interact with matter.
4 0
3 years ago
A radioactive particle has a half life of 1 second. If it moves at 3/5 the speed of light, from my point of view standing still
AVprozaik [17]

The half-life observed by an observer at rest is 1.25 s

Explanation:

In this problem, the particle is travelling at a significant fraction of the speed of light: therefore, we have to consider the phenomenon of the time dilation. The half-life of the particle as measured by an observer at rest is given by

T=\frac{T_0}{\sqrt{1-(\frac{v}{c})^2}}

where

T_0 is the proper half-life of the particle

v is the speed of the particle

c is the speed of light

For the particle in this problem,

T_0 = 1 s

v=\frac{3}{5}c=0.6 c

Substituting into the equation, we find:

T=\frac{1}{\sqrt{1-(\frac{0.6c}{c})^2}}=1.25 s

#LearnwithBrainly

7 0
3 years ago
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