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3 years ago

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An object with a 25 µC charge is 0.54 m away from a second charged object. They experience a force of 3250 N. What is the charge

on the second object?
Your answer

1 answer:

nekit [7.7K]3 years ago

6 0

25 uc charge is 0.54 m away

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A system absorbed 44 joules of heat from its surroundings. After doing work, the increase in the internal energy of the system i

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Idk sorry maybe try @amyletbe

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A brick of mass 2.0kg is at rest. It falls to the ground through a

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Answer:

I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

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3 years ago

Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha

leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q / ε₀

E = Q / 4πε₀r²

E ∝ 1 / r² .

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3 years ago

There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in

qaws [65]

Answer:

Pressure increases due to enlargement

Explanation:

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Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

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from the next 2, we get that since d2<d1, AP3>AP2

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for fluids flowing in tubes (blood vessel in this case)

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masha68 [24]

Answer:

The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]

Explanation:

We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.

For this case, we don't use the elevation difference and therefore those terms can be cancelled.

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