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adelina 88 [10]
2 years ago
6

Three cars collide, a 1500 kg sports car, a 1750 kg family car, and a 1200 kg compact car. Which experiences the greatest change

in momentum?
a. the sports car
b. the family car
c. the compact car
d. not enough information
e. all the same
Physics
1 answer:
sweet [91]2 years ago
5 0

Answer:

d

Explanation:

This is because momentum is defined as p = mv

delta p = Force *time

neither velocity nor time is given so a conclusion cannot be made on which has the greatest momentum change.

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Answer:

Velocity

Explanation:

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The set of frequencies of the electromagnetic waves emitted by the atoms of an element is called
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7 0
2 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
ANEK [815]

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

6 0
3 years ago
Read 2 more answers
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