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maw [93]
3 years ago
12

You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the

gas pedal. You can ignore friction, but you can't ignore air resistance. Identify all the forces acting on the car?
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

Weight, air drag, normal.

Explanation:

Since the car is only rolling downhill, the forces that will come into play are: it's own weight (the gravitational force the Earth exerts upon the car), the air resistance (the force the air exerts upon the car, since it says it can't be ignored) and finally the normal force (the contact force the ground exerts on the car). There are no other forces sources.

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1 year ago
Which of the following is not an example of centripetal acceleration?
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An apple falling to the ground is not an example of centripetal acceleration.
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3 years ago
G to ml,density=3.291 g/ml
olya-2409 [2.1K]
The density value 3.291 g/ml (gram / milliliter) in words is "three point two nine one g/ml (gram / milliliter)".
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The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha
denis-greek [22]

Answer:

H=41.3kJmol^{-1}

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}

if we insert values, we arrive at

ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}

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3 years ago
Inceptual Physics - Wavos Test - Semester 1 - 2021
Dvinal [7]

Answer:

A thats answer thankyou

4 0
2 years ago
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