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scZoUnD [109]
2 years ago
5

The number of gallons of water in a storage tank at time t, in minutes, is modeled by w(t) = 25 − t2 for 0≤t≤5. At what rate, in

gallons per minute, is the amount of water in the tank changing at time t = 3 minutes?
Physics
1 answer:
Kisachek [45]2 years ago
8 0

Answer:

Rate of change of water will be -6 gallon/minute

Explanation:

We have given water in the tank as the function of time as

w(t)=25-t^2

We have to find the rate of change of water in the tank at t = 3 minute

For rate of change we have to differentiate both side

So \frac{dw}{dt}=0-2t

At t = 3 minute

\frac{dw}{dt}=0-2\times 3=-6gallon/minute

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west 1100 km / hr    ..

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I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

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2 years ago
A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

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r is the distance from the centre of the sphere

Here we have:

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Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

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q is the charge on the plastic cube

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m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

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Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
2 years ago
How is earth outer layer different from a cracked hard-boiled egg?
Aleksandr [31]

Answer:

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Replacing at the expression to frequency of observer we have,

f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})

f_{obs} = 878Hz

Therefore the frequency receive by observer is 878Hz

8 0
2 years ago
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