NEED HELP

Jon's bathtub is rectangular and its base is 18 ft2. (a) How fast is the water level rising if Jon is filling the tub at a rate

Kitty [74]

Answer:

The water level in the bath tub is rising at a rate of 0.0111 ft/s

Explanation:

Volume of the bath tub = (Area of base) × (height)

Area of base = 18 ft² (constant)

Height = h (variable)

V = 18h

(dV/dt) = 18 (dh/dt)

If (dV/dt) = 0.2 ft³/s

0.2 = 18 (dh/dt)

(dh/dt) = (0.2/18)

(dh/dt) = 0.0111 ft/s

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

If you want to conduct an electrical current, which situation would produce a solution capable of this? A) Dissolving water in o

svp [43]

D dissolving solid NaF in water

6 0

3 years ago

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Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE

GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0

3 years ago

How does the amount of atmosphere affect what color the sky appears to be?

weeeeeb [17]

The atmosphere creates a cover over the earth preventing us from seeing space and it’s natural color and the light that is created by the sun reflected on the ocean creates the color that the sky appears to be blue

3 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago