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kipiarov [429]
3 years ago
13

A man paddles a canoe at 6 km per hour. If he paddles on a river with a current of 6 km per hour, what is the speed of the canoe

if it heads: Upstream? Directly across the river?

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is  8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]

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Momentum = mass*velocity(or speed)

241.5 kg*m/s = 115 kg * v
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3 years ago
The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
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Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

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Area = F / P
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Hope this answers the question. Have a nice day.
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