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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The frequency for a fundamental pipe is given as:
f = v/4L
L is equal to the length of the pipe
Since L = Lo/2 where Lo is the original length of the pipe, the
new frequency would be:
f = (v/4)/(Lo/2)
f = 2 (v/4Lo)
Since v/4Lo = fo, therefore:
f = 2 fo
Answer:
v = 4.10 10⁻³ m / s
Explanation:
For this exercise we will use Newton's second law where the force is magnetic
F -W = m a
As the field is directed to the north and the proton to the east, using the rule of the right hand the force is vertical upwards, the force balances the weight the acceleration is zero
F = W
q v B = m g
Let's calculate the speed
v = m g / (q B)
v = 1,673 10⁻²⁷ 9.8 / (1.6 10⁻¹⁹ 2.5 10⁻⁵)
v = 4.10 10⁻³ m / s
Answer:
-39.2m/s
Explanation:
Given that :
t = 4secs
g = -9.8m/s^2
v = ?
u = 0m/s ( since it was at rest )
V = u +at............. 1
Where v is the final velocity
a = -g = -9.8m/s^2 since the ball was dropped from a height which will eventually make it move against gravity
t = 4secs
Substitute the values into 1
v = 0 - 9.8×4
v = -39.2m/s
Answer:
216 m
Explanation:
Assuming a straight line:
Δx = vt
Δx = (12 m/s) (18 s)
Δx = 216 m
