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Ilya [14]
2 years ago
8

A water tank is in the shape of an inverted cone with depth 10 meters, and top radius 8 meters. Water is flowing into the tank a

t a rate of 0.1 cubic meters / min, but leaking out at a rate of 0.002h^20.002 h 2 cubic meters / min, where hh is the depth of water in the tank, in meters. Find the depth of water when the volume of water in the tank is neither increasing nor decreasing.
Physics
1 answer:
Elis [28]2 years ago
7 0

Answer:

The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.

The depth of the water has to be equal to 7.07 m in order to have a stationary volume.

Explanation:

In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:

0.002*h^2 = 0.1

h^2 = 0.1/0.002

h^2 = 50

h = sqrt(50) = 7.07 m

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Answer:

0.2

Explanation:

3/15

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A race car has a mass of 820 kg. It starts from rest and travels 50.0m in 3.0s. The car is uniformly accelerated during the enti
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A=DELTAv/DELTAt=50/3
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2 years ago
A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides
lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

\Delta P = P_f - P_i

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

Learn more about momentum here: brainly.com/question/7538238

8 0
2 years ago
1. A runner drops her phone as she is running at a constant speed of 3 miles per hour from point A to point B in a park. Describ
joja [24]

Answer:

Let's define the point A as our zero in the x-axis.

As the phone drops, it keeps the horizontal velocity that it had before, so the horizontal velocity is:

Vx = 3 mi/h.

Now, the only force acting on the phone is the gravitational force that acts in the vertical axis, then we have:

Ay = -g

where g = 9.8 m/s^2

It is dropped, so we do not have a vertical initial velocity, then for the vertical velocity we should integrate over time:

Vy = -g*t

And for the position again, we integrate over time, but now we have an initial position H, that is the height at which the phone is dropped.

Py = -(1/2)*g*t^2 + H

And the horizontal position can be found by integrating over time the horizontal velocity.

Px = (3mi/h)*t

This will be the two equations that describe the motion of the phone, and we can not solve it further because we do not know the initial height of the phone.

But in general, we have a linear equation in the horizontal axis and a quadratic equation with a negative leading coefficient in the vertical axis.

Position(t) = ( (3mi/h)*t,  -(1/2)*g*t^2 + H)

5 0
3 years ago
A rock is thrown horizontally from a bridge with a velocity of 17.0 m/s. It takes the rock 3.0 s to strike the water below. What
zalisa [80]

Answer:

V= 33.98 m/s

Explanation:

Given that

Horizontal speed ,u= 17 m/s

Time taken by rockets to strike the water ,t= 3 s

We know that acceleration due to gravity ,g= 9.81 m/s²

There is no any acceleration in the horizontal direction that is why the horizontal veloity will remain constant.

In the vertical direction

vy = uy+ g t

Initial velocity in vertical direction is 0 m/s.

vy= 0+ 9.81 x 3

vy = 29.43 m/s

The resultant velocity

V=\sqrt{v_y^2+u^2}

V=\sqrt{29.43^2+17^2}\ m/s

V= 33.98 m/s

7 0
3 years ago
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