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3 years ago

The atmospheric pressure in denver, co is 633 mmhg. what is this pressure in atm?

1 answer:

7 0

1 atm of pressure corresponds to 760 mmHg. Therefore, we can set a simple proportion to find how many atmospheres of pressure correspond to 633 mmHg, in the following way:
$1 atm : 760 mmHg = x : 633 mmHg$
And if we solve this proportion, we find the pressure in atmospheres:
$x= \frac{1 atm \cdot 633 mmHg}{760 mmHg}=0.83 atm$

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A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

3 years ago

A tire has a pressure of 325 kPa at 10°C.

musickatia [10]

5 times that of initial pressure i.e 1625 kpa

6 0

3 years ago

Read 2 more answers

A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai

Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a) Let us assume the depth be h

As we know that

$Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h$

After solving this,

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

Given that

height of the extra fluid is

$h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}$

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

$Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000$

Pb' = 122 KPa

3 0

3 years ago

According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of wate

WARRIOR [948]

Answer:

Explanation:

Given:

1 can of concentrate requires 3 cans of water

Now,

Total ounces in 200 6-ounce cans = 1200 ounces

also,

for 1 can of concentrate requires 3 cans of water

thus,

for 12 ounces can water can required = 3 × 12 ounces = 36 ounces of cans

Thus,

total ounce of juice per can = 12 + 36 = 48 ounces per can

therefore,

the number of 12-ounce cans required are = $\frac{\textup{Ttoal ounces}}{\textup{Ounces per can}}$

= $\frac{\textup{1200}}{\textup{48}}$

the number of 12-ounce cans required are = 25

8 0

3 years ago

One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and

taurus [48]

Answer:

(a) Initial volume will be 7.62 L

(b) Final temperature will be 303.85 K

Explanation:

We have given one mole of ideal gas done 3000 J

So work done W = 3000 J

Let initial volume is $V_1$ and initial pressure $P_1=1atm$ ( As pressure is constant )

Final volume $V_2=25L$ = 0.025 $m^3$

Number of moles n = 1

(B) From ideal gas of equation we know that $PV=nRT$

So $1.01\times 10^5\times0.025=1\times 8.31\times T$

T = 303.85 Kelvin

(B) For isothermal process work done is equal to

$W=nRTln\frac{V_2}{V_1}$

$3000=1\times 8.314\times 303.85\times ln\frac{0.025}{V_1}$

$ln\frac{0.025}{V_1}=1.1881$

$\frac{0.025}{V_1}=3.2808$

$V_2=0.00846m^3=7.62L$

So initial volume will be 7.62 L

5 0

3 years ago