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Aleks04 [339]
3 years ago
5

A spherical balloon is filled with gas at a rate of 4 cm 3 /s . at what rate is the radius r changing with with respect to time

when the volume v = 36 π cm 3 ?
Physics
1 answer:
Gnoma [55]3 years ago
8 0
The volume of the sphere is given by V = 4/3 * pie * r^3. We seek dr/dt that the rate of change of the radius with respect to time.  
V = 4/3 * pie * r^3 
Since we know the rate at which the volume changes wrt time. We can plug it in to find dr/dt.
 dV/dt = 36pie. But dV/dt = 4/3 * pie * 3r^2 * dr/dt
 So we have that dr/dt = (dV/dt) / 4/3 * pie * 3r^2
 So dr/dt = (36pie)/ 4/3 * pie * 3(4)^2
 dr/dt = 113.112/ 201.0864 = 0.5624

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In which one or more of the following is the earth assumed to be the center of the universe?
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4 0
3 years ago
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A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

            Force =  Spring constant x Elongation

            F = kx

Here force is weight of mass

           F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x  = 36 cm = 0.36 m

Substituting

           F = kx

           5.3 = k x 0.36

             k = 14.72 N/m

Spring constant is 14.72 N/m

6 0
3 years ago
A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
NARA [144]
To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters
8 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
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How much time is needed to produce 720 Joules of work if 90 watts of power is used?
Tems11 [23]

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

power is measured in Watts (W) , when the work done is measured in Joules (J) and time in seconds

by the definition of power,

Power=\frac{work.done}{time.taken} \\ \\ time.taken=\frac{work.done}{power}\\=\frac{720J }{90W} \\ \\ =8 s

3 0
3 years ago
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