To solve this problem it is necessary to apply the concepts related to the conservation of energy, specifically the potential elastic energy against the kinetic energy of the body.

By definition this could be described as

$PE = KE$

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

Where

k = Spring constant

x = Displacement

m = mass

v = Velocity

This point is basically telling us that all the energy in charge of compressing the spring is transformed into the energy that allows the 'impulse' seen in terms of body speed.

If we rearrange the equation to find v we have

$v = \sqrt{\frac{kx^2}{m}}$

Our values are given as

$m = 1000kg$

$k = 5.75*10^6N/m$

$x = 3.12*10^{-2}m$

Replacing at our equation we have then,

$v = \sqrt{\frac{kx^2}{m}}$

$v = \sqrt{\frac{(5.75*10^6)(3.12*10^{-2})^2}{1000}}$

$v = 2.3658m/s$

Therefore he speed of the car before impact, assuming no energy is lost in the collision with the wall is 2.37m/s

4 0

3 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

More pts first 2 so hurry up

Norma-Jean [14]

Answer:

oh 50 points! how did you do it??!?!?!?! I see up to 8 points only

8 0

3 years ago

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When analyzing a position time graph the direction of the slope positive or negative tells us:

Sphinxa [80]

Answer:

Explanation:

Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.

8 0

3 years ago

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