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1 year ago

9

A stable orbit is an orbit that repeats indefinitely, without ever changing shape. After running several simulations with differ

ing parameters, what attributes did all the stable orbits have in common?

1 answer:

Tresset [83]1 year ago

3 0

Answer:

All of the orbits were in the shape of an ellipse, with the orbited body on the inside of the ellipse.

Explanation:

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Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago

Lucille is finding it difficult to play soccer after school. Her doctor thinks that her cells might not be getting enough oxygen

Bad White [126]

Explanation:

Lucile

Lucile cells needs a huge abundance of glucose and oxygen supply in order for her body to function properly and participate in a soccer game.

Playing a soccer game is energy demanding. The body's energy supply is met by the metabolism of glucose in cells using oxygen.

The problem with Lucille's respiratory and circulatory system can make it difficult for her to take part in soccer game because the needed energy would not be available for her to participate in the game.

Respiratory system is saddled with the responsibility of bringing enough oxygen to body system through gaseous exchange and metabolism.
The circulatory system carries the oxygen to the different body cells to where they are need. It also takes the deoxygenated blood to where they can get oxygenated.
We can clearly see that if respiratory and circulatory systems are not functioning well, Lucille's cells will not be able to furnish her with the much needed energy.

Ezra

Ezra needs energy to play basketball well. The energy is produced via the glucose in the starch and the oxygen from breathing.

When starch is broken down, glucose is produced.

How it works;

The digestive, circulatory and respiratory systems are all important here. They ensure that Ezra gets the needed energy from the food he is eating. Proteins also have the capability of producing energy when there is shortage of starch in the body.

Ezra's digestive system through a couple of processes releases glucose from the food he eats especially the bread in the sandwich.

The glucose is stored in the body and made available for use when in need.

The glucose is used to produce energy.

When Ezra breathes, the circulatory system ensures gaseous exchange with the surroundings. Oxygen rich blood is taken the cells where they combine with glucose for their metabolic actions. Oxygen deficient bloods are carried away.

In the cell, a special apparatus called the mitochondria combines glucose with oxygen and energy is liberated. The energy is used by Ezra to run while playing basketball

learn more:

Cellular respiration brainly.com/question/3447259

#learnwithBrainly

7 0

2 years ago

You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By

Fittoniya [83]

Answer:

The frequency is $f = 0.221 \ Hz$

Explanation:

From the question we are told that

The time taken for it to decay to half its original size is $t = 3.40 \ ms = 3.40 *10^{-3} \ s$

Let the voltage of the capacitor when it is fully charged be $V_o$

Then the voltage of the capacitor at time t is said to be $V = \frac{V_o}{2}$

Now this voltage can be mathematical represented as

$V = V_o * e ^{-\frac{t}{RC} }$

Where RC is the time constant

substituting values

$\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }$

$0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }$

$- \frac{0.5}{RC} = ln (0.5)$

$-\frac{0.5}{RC} = -0.6931$

$RC = 0.721$

Generally the cross-over frequency for a low pass filter is mathematically represented as

$f = \frac{1}{2 \pi * RC }$

substituting values

$f = \frac{1}{2* 3.142 * 0.72 }$

$f = 0.221 \ Hz$

7 0

1 year ago

Come si compongono due forze che agiscono in diversi punti di un corpo rigido? Oof

bagirrra123 [75]

Answer:

Explanation:

I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

7 0

1 year 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.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
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

1 year ago