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3 years ago

5

At an ocean depth of 10.0m, a diver's lung capacity is 2.40L. The air temperature is 32.0°C and the pressure is 101.30 kPa. What

is the volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa?

1 answer:

Inessa [10]3 years ago

8 0

Answer: The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 101.30 kPa

$P_2$ = final pressure of gas = 141.20 kPa

$V_1$ = initial volume of gas = 2.40 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $32.0^0C=(32+273)K=305K$

$T_2$ = final temperature of gas = $21.0^0C=(21+273)K=294K$

Now put all the given values in the above equation, we get:

$\frac{101.30\times 2.40}{305}=\frac{141.20\times V_2}{294}$

$V_2=1.66L$

The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L

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A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what

Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

$v^2 = u^2 + 2as$

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

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Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0 as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

$v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25$

Thus, initial speed with the truck was travelling was 25 m/s.

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3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

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Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

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Alexandra is attempting to drag her 32.6 kg golden retriever across the wooden floor by applying a horizontal force. What force

vichka [17]

Answer:

230.26 N

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3 years ago

A net force of 60 N north acts on an object with a mass of 30 kg. Use Newton's second law of

earnstyle [38]

Answer:

Explanation:

F = ma. For us, this looks like

60 = 30a and

a = 2 m/s/s

If the force goes up to, say, 90, then

90 = 30a and

a = 3...if the force goes up, the acceleration also goes up.

If the mass goes up to say, 60, and the force stays the same, then

60 = 60a and

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3 years ago

A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

or

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

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on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

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n = 0.563

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b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

or

Average kinetic energy = 6.04 × 10⁻²¹ J

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or

rms speed = $\frac{3\times8.314\times292}{0.004}$

or

rms speed = 1349.35 m/s

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3 years ago