A student is standing 8 m from a roaring truck engine that is measured at 20 dB. The student moves 4 m closer to

a glass bottle of soda is sealed with a screw cap the absolute pressure of the gas inside the bottle is 1.80*10^5 pa. assuming t

eduard

The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).

We can multiply both sides of the formula by S to obtain the force.

P*S=(F*S)/S

P*S=F

Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])

(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F

73.8 N =F

8 0

2 years ago

A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.

zzz [600]

Answer:

0.0257259766982 m

Explanation:

$P_2$ = Atmospheric pressure = 101325 Pa

$d_1$ = Initial diameter = 1.5 cm

$d_2$ = Final diameter

$\rho$ = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

$P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa$

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m$

The diameter of the bubble is 0.0257259766982 m

8 0

2 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$