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3 years ago

11

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

to be 296.0 mm Hg. At what temperature (in °C) will water boil at base camp ? The vapor pressure of water at 373 K is 760.0 mm Hg. (ΔH°vap for H2O = 40.7 kJ/mol and R = 8.314 J/mol K)

1 answer:

Mamont248 [21]3 years ago

4 0

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

$ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K$

T₁ = 347.996 K = 74.86°C

∴Water will boil at 74.86°C

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Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22

natka813 [3]

Answer:

2420 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 22.9 N

Angle (θ) = 35°

Distance (d) = 129 m

Workdone (Wd) =?

The work done can be obtained by using the following formula:

Wd = Fd × Cos θ

Wd = 22.9 × 129 × Cos 35

Wd = 22.9 × 129 × 0.8192

Wd ≈ 2420 J

Thus, the workdone is 2420 J.

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3 years ago

A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?

klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

6 0

3 years ago

Read 2 more answers

Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0

2 years ago

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$

Then, plugging in the given values, we obtain:

$k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m$

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0

3 years ago

Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

Nitella [24]

Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos:

$T = \frac{1}{f}$ (1)

Donde $f$ es la frecuencia, en hertz.

( $f = 4\,hz$ )

$T = \frac{1}{4\,hz}$

$T = 0.25\,s$

Ahora determinamos la fuerza aplicada sobre la prenda ( $F$ ), en newtons:

$F = m\cdot a$ (2)

$F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}$ (2b)

Donde:

$m$ - Masa de la prenda, en kilogramos.

$r$ - Radio interior del tambor, en metros.

( $m = 0.32\,kg$ , $r = 0.4\,m$ , $T = 0.25\,s$ )

$F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}$

$F \approx 80.852\,N$

La velocidad lineal de la lavadora es:

$v = \frac{2\pi\cdot r}{T}$ (3)

( $r = 0.4\,m$ , $T = 0.25\,s$ )

$v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}$

$v \approx 10.053\,\frac{m}{s}$

Y la velocidad angular del tambor de la lavadora:

$\omega = \frac{2\pi}{T}$

( $T = 0.25\,s$ )

$\omega = \frac{2\pi}{0.25\,s}$

$\omega \approx 25.133\,\frac{rad}{s}$

7 0

2 years ago