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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Explanation:
I hesitate to articulate upon the highest degree of accuracy!
The only real difference is that common seismic waves travel through the ground and sound waves travel through the air. If you had a pipe attached to granite and you were listening to it, you might detect both.
For a given closed circuit the current voltage and resistance of circuit is related with ohm's law

here
V = voltage difference due to battery
i = current flow in the circuit
R = resistance
so from above equation

so in order to decrease the current in the circuit we can either increase the resistance of the circuit or decrease the voltage
Now the formula of resistance is

here
= resistivity of material of resistance

A = crossectional area
a) if we use a thick wire the as per above formula of resistance the value of resistance will decrease.
And hence the current will increase in the circuit so it is not correct option
b) If 6V battery is replaced by 12 V battery then due to increase in potential difference the current in the circuit will also increase, so this is not correct answer.
c)Opening the switch will decrease the voltage from 6 V to 0 Volt
due to this decrease in potential difference the current in the circuit will also decrease. So by Opening the switch the current will be decreased in this circuit.
Answer:
The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>
Explanation:
Given:
Upward direction is positive. So, downward direction is negative.
Tota time the ball remains in air (t) = 8.0 s
Net displacement of the ball (S) = Final position - Initial position = 0 m
Acceleration of the ball is due to gravity. So,
(Acting down)
Now, let the initial velocity be 'u' m/s.
From Newton's equation of motion, we have:

Plug in the given values and solve for 'u'. This gives,

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.