0.119cm/s is the radius of the balloon increasing when the diameter is 20 cm.
<h3>How big is a circle's radius?</h3>
The radius of a circle is the distance a circle's center from any point along its circumference. Usually, "R" or "r" is used to indicate it.
A circle's diameter cuts through the center and extends from edge to edge, in contrast to a circle's radius, which extends from center to edge. Essentially, a circle is divided in half by its diameter.
dv/dt = 150cm³/s
d = 2r = 20cm
r = 10cm
find dr/dt
Given that the volume of a sphere is calculated using
v = 4/3πr³
Consider both sides of a derivative
d/dt(v) = d/dt( 4/3πr³)
dv/dt = 4/3π(3r²)dr/dt = 4πr²dr/dt
Hence,
dr/dt = 1/4πr².dv/dt
dr/dt = 1/4π×(10)²×150
dr/dt = 1/4π×100×150
dr/dt = 0.119cm/s.
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Answer:
C. I think
Explanation:
C. permanent positive charges
Answer:
*want my husband's mouth to stop moving
*when im driving my car and the light turns red
*i want the person pushing the buggy behind me to stop when I stop
*when I'm on the 16th floor at the Horseshoe Casino Hotel and the elevator keeps passing my floor... would love for it to stop and pick me up
*when Publisher's Clearing House is delivering my big fat check, I definitely want them to stop and hand it over:)
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
Answer:
24.9 m
Explanation:
A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.
You walk 21.5 m straight east and then 12.5 m straight north. This forms a right angled triangle with a horizontal distance of 21.5 m, a vertical distance of 12.5 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Therefore using Pythagoras theorem:
x² = 21.5² + 12.5²
x² = 462.25 + 156.25
x² = 618.5
Taking square root of both sides:
√x² = √618.5
x = √618.5
x = 24.9 m = distance between the ending and starting point.
