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Julli [10]
2 years ago
9

List the jovian planets in order of increasing distance from the sun?

Physics
1 answer:
trapecia [35]2 years ago
5 0
Here are the planets listed in order of their distance from the Sun: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. An easy mnemonic for remembering the order is “My Very Educated Mother Just Served Us Noodles

Hope this helps.
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2 rocks of the same mass are rolling down a hill at different speeds (one faster than the other). which rick would have more kin
stepan [7]

Kinetic energy of an object depends on its mass and its speed.

If the masses are equal, then the rick with <em>more speed</em> has more kinetic energy.

5 0
3 years ago
Read 2 more answers
Can someone please explain what inertia is?
Bad White [126]
If an object is in motion it will stay in motion 
8 0
2 years ago
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What would be the best way for her to do this?
andrew-mc [135]
Need more info plz :)


5 0
3 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
3 years ago
The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I
erma4kov [3.2K]

Answer:

Part a)

I_{end} = \frac{mL^2}{3}

Part b)

I_{edge} = \frac{2ma^2}{3}

Explanation:

As we know that by parallel axis theorem we will have

I_p = I_{cm} + Md^2

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

I = \frac{mL^2}{12}

now if we need to find the inertia from its end then we will have

I_{end} = I_{cm} + Md^2

I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2

I_{end} = \frac{mL^2}{3}

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

I = \frac{ma^2}{6}

now if we need to find the inertia about an axis passing through its edge

I_{edge} = I_{cm} + Md^2

I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2

I_{edge} = \frac{2ma^2}{3}

7 0
3 years ago
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