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vlada-n [284]
3 years ago
4

Jeff has 10 grams of water and 10 grams of vegetable oil in separate containers. Both liquids have a temperature of 24°C. Jeff h

eats both liquids over a flame for five minutes. When he’s finished, he discovers that the temperature of the oil increased more than the temperature of the water. What can Jeff conclude from this experiment?
Physics
2 answers:
jekas [21]3 years ago
0 0
Olive oil<span> </span>heated<span> up </span>faster than water <span>because the </span>heat<span> capacity of </span>oil<span> is lower </span>than<span> the </span>heat<span> capacity of </span>water<span>. </span>Water <span>requires more energy per gram of liquid to change its temperature.</span>
Reptile [31]3 years ago
0 0
Oil heats hotter and faster than water due to its components

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svet-max [94.6K]

Answer:

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

S = ut + \frac{1}{2}at^{2}  \\

Where S is distance traveled

u is the initial velocity

t is time

and a is acceleration

Since the train starts from rest, u = 0 m/s

Hence,  for the first 20s

a = 1.1 m/s²; t = 20s,  u = 0 m/s

∴ S = ut + \frac{1}{2}at^{2}  \\ gives

S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}

S =  \frac{1}{2}(1.1)(20)^{2}

S = 220m

This is the distance covered in the first 20s.

  • The train then proceeds at constant speed for 1100m.

Now, we will calculate the speed attained here

From

v = u +at

Where v is the final velocity

Hence,

v = 0 + 1.1(20)

v = 1.1(20)

v = 22 m/s

This is the constant speed attained when it proceeds for 1100m

  • The train then slows down at a rate of 2.2 m/s² until it stops

We can calculate the distance covered while slowing down from

v^{2} = u^{2} + 2as

The initial velocity, u here will be the final velocity before it started slowing down

∴u = 22 m/s

The final velocity will be 0, since it came to a stop.

∴ v = 0 m/s

a = -2.2 m/s² ( - indicates deceleration)

Hence,

v^{2} = u^{2} + 2as gives

0^{2} =22^{2} +2(-2.2)s

0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m

This is the distance traveled while slowing down.

A) The distance between the stations is

220m + 1100m + 110m

= 1430m

Hence, the distance between the stations is 1430m

B) The time it takes the train to go between the stations

The time spent while accelerating at 1.1 m/s² is 20s

We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

From,

Speed =\frac{Distance}{Time}\\

Then,

Time = \frac{Distance}{Speed}

Time = \frac{1100}{22}

Time = 50 s

And then, the time spent while decelerating (that is, while slowing down)

From,

v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s

This is the time spent while slowing down until it stops at the station.

Hence, The time it takes the train to go between the stations is

20s + 50s + 10s = 80s

The time it takes the train to go between the stations is 80s

6 0
3 years ago
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