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Citrus2011 [14]
3 years ago
15

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 5 cubic feet

per minute. If the pool has radius 6 feet and height 9 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet
Physics
1 answer:
Marina86 [1]3 years ago
7 0

Answer:

\frac{dH}{dt}=0.044\;\;feet\;per\;min

Explanation:

Given,

Radius R = 6 feet

Height H = 9 feet

Depth of the water = 5 feet

Volume of right circular cylinder,

V=\pi R^2 H\\

Differentiate with respect to the H

\frac{dV}{dH}=\pi R^2\\\frac{dV}{dH}=36\pi

Now,

\frac{dV}{dH}.\frac{dt}{dt} =36\pi\\\frac{dV}{dt}.\frac{dt}{dH}=36\pi\\  5.\frac{dt}{dH}=36\pi\\\frac{dH}{dt}=\frac{5}{36\pi}\\\frac{dH}{dt}=0.044\;\;feet\;per\;min

Hence, the rate of chage of height of water is 0.044 feet per min.

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Answer:

|F| = 393750  N

Explanation:

Given that,

Total mass of the train, m = 750000 kg

Initial speed, u = 84 m/s

Final speed, v = 42 m/s

Time, t = 80 s

We need to find the net force acting on the train. The formula for force is given by :

F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{750000\times (42-84)}{80}\\\\F=-393750\ N

So, the magnitude of net force is 393750  N.

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give an example of situation in which an automobile driver can have a centripetal acceleration but no tangential speed
Mars2501 [29]

There is no need for tangential acceleration when moving in a circle at a constant speed.

<h3>What is centripetal acceleration?</h3>

centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.

<h3>Which is an example of centripetal acceleration?</h3>

Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.

To know more about tangential acceleration :

brainly.com/question/14993737

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1 year ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
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A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

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Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

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The potential energy of a watermelon is 15.0 J. The watermelon is 3.0 m high. What is the mass of the watermelon?
maks197457 [2]

Answer:

m = 0.51[kg]

Explanation:

Potential energy is defined as the product of mass by gravity by height.

E_{pot}=m*g*h

where:

Epot = potential energy = 15 [J]

m = mass [kg]

g = gravity acceleration = 9.8 [m/s²]

h = elevation = 3 [m]

Now replacing:

E_{pot}=m*g*h\\15=m*9.8*3\\m = 0.51[kg]

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3 years ago
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