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Nostrana [21]
3 years ago
15

Suppose for an unexplained reason a star near Earth started shrinking. At what stage is the gravitational force between Earth an

d the star the greatest? Stage 1 Stage 2 Stage 3 Stage 4
Physics
2 answers:
tatiyna3 years ago
3 0

the answer is A- stage 1

DENIUS [597]3 years ago
3 0

stage 1 is the correct answer!

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Two charges, - Q0 and - 4Q0 are a distance d apart. These two charges are free to move but do not because there is a third charg
TEA [102]

To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

q_1 = -Q_0

q_2 = -4Q_0

As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be q_3 = +Q is placed at a distance x from q_1

F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}

The force on q_3 due to q_2 is

F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}

The condition of equilibrium is

F_{1,3} = F_{2,3}

\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}

\frac{1}{x^2} = \frac{4}{(1-x)^2}

x = 0.331 from q_1

To find the magnitude of q_3 we use F_{1,2} = F_{1,3}

\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}

Q = 0.43Q_0

The magnitude of the third charge must be 0.43 the first charge Q_0

3 0
4 years ago
A mixture of helium and argon gas is expanded from a volume of 29.0L to a volume of 82.0L, while the pressure is held constant a
finlep [7]

Answer:

322 kJ

Explanation:

The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.

When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.

For a constant pressure, the work can be calcutated by:

W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:

p = 60 atm = 6.08x10⁶ Pa

ΔV = 82.0 - 29.0 = 53 L = 0.053 m³

W = 6.08x10⁶x0.053

W = 322x10³ J

W = 322 kJ

8 0
3 years ago
Water flows horizontally through a garden hose with an inner diameter of .012 m at a speed of 7.8 m/s. it exits out a small nozz
tigry1 [53]

<span>Answer: 110.12 m/s </span>

We will use the formula A1V1 = A2V2 where 7.8 m/s is divided with 0.0085 m then multiply to 0.12 m, the result will be 110.117 or 110.12 m/s. This is related to the continuity of fluid flow in which as liquid moves horizontally, the same amount of liquid goes out as it comes in or the liquid itself do not change as it moves but the speed does when the diameter changes.

 

3 0
3 years ago
Read 2 more answers
An object at rest increases its velocity to 10 m/s in 3 seconds. what is its acceleration?
Anna11 [10]

Answer:

30m

Explanation:

10m per second

there is 3 seconds

10 times 3 is 30

4 0
4 years ago
Which of the following is characteristic of the second stage within a demographic transition?
Elan Coil [88]
The answer is A) <span>The death rate begins to fall, but birth rates remain high for a time.</span>
3 0
4 years ago
Read 2 more answers
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