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Elena-2011 [213]
1 year ago
11

10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball

Physics
1 answer:
SVEN [57.7K]1 year ago
3 0

Answer:

Josie's ball faster than T

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Write in the word "stronger" or "weaker." The bigger and
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Answer:

weaker has the heavier of an object

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3 years ago
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the
pantera1 [17]

Answer:

Explanation:

F = k*n

F = 30N

k = ???

N = 0.73 M

F = k* N

30N = k * 0.73

k = 30N / 0.73

k = 41.1

5 0
2 years ago
Which example describes a nonrenewable resource?
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The refineries that use the oil to put in their cars as gasoline and then after a while the oil will disappear and go away and that's what a nonrenewable resource would be
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Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0
2 years ago
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}

4 0
3 years ago
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