4.

A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each

jekas [21]

Answer:

884 balloons

Explanation:

Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.

So if pressures reduces from 100 to 1.2, the new volume would be

$V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2$

The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is

$V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3$

The number of balloons that 12.5 m3 can fill in is

$V_2/V_b = 12.5 / 0.014 = 884$

8 0

3 years ago

A playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the gro

Shtirlitz [24]

Answer:

1.94601 rad/s

Explanation:

$I_1$ = Moment of inertia of carousel = 124 kgm²

$\omega_1$ = Angular speed of carousel = 3.5 rad/s

$\omega_2$ = Angular speed of person

r = Radius of carousel = 1.53 m

m = Mass of person = 42.3 kg

Moment of inertia of person

$I_2=mr^2\\\Rightarrow I_2=42.3\times 1.53^2\\\Rightarrow I_2=99.02007\ kgm^2$

As the angular momentum is conserved in the system

$I_1\omega_1=(I_1+I_2)\omega_2\\\Rightarrow \omega=\frac{I_1\omega_1}{(I_1+I_2)}\\\Rightarrow \omega_2=\frac{124\times 3.5}{(124+99.02007)}\\\Rightarrow \omega_2=1.94601\ rad/s$

The angular speed of the carousel after the person climbs aboard is 1.94601 rad/s

5 0

3 years ago

A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the

Elina [12.6K]

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

T = $2\pi \sqrt{ \frac{L}{g} }$

T² = $4\pi ^2 \frac{L}{g}$ 4pi2 L / g

g = $4\pi ^2 \frac{L}{T^2}$

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

g = $4\pi ^2 \frac{0.823}{1.823^2}$ 4 pi2 0.823 / 1.823 2

g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

T = t / n

ΔT = $\frac{dT}{dt}$ Δt + $\frac{dT}{dn}$ ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

ΔT = Δt / n

ΔT = Δt

now let's look for the uncertainty of g

Δg = $\frac{dg}{dL}$ ΔL + $\frac{dg}{dT}$ ΔT