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schepotkina [342]
3 years ago
6

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

of 0.2 ft3/min? (Round your answer to three decimal places.)

Physics
2 answers:
Kitty [74]3 years ago
6 0

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!!!

gizmo_the_mogwai [7]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

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