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aniked [119]
3 years ago
9

Dalton’s completing an investigation in the science lab. He observes that a sample of liquid turns to gas at 135°C. What’s this

temperature called? A. boiling point B. freezing point C. melting point D. room temperature E. standard temperature
Physics
2 answers:
FinnZ [79.3K]3 years ago
4 0

Answer:

option a

Explanation:

artcher [175]3 years ago
3 0

The temperature at which the sample of liquid turns to gas at 135 °C is termed as boiling point.

Answer: Option A

<u>Explanation: </u>

The observation of conversion of liquid to gas indicates that there is occurrence of change in the state of matter. The inter-conversion from one state to another can be done by either varying the temperature or by varying the pressure.

In this case, the liquid on heating gets converted to gaseous state after attaining a particular temperature say 135 °C. So, this process of conversion from liquid to gaseous state on heating is termed as boiling.

The temperature at which a liquid converts to gas is termed as the boiling point of that liquid.

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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
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<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>

<span>t = √(2y/g) </span>

<span>in the ft - lb - s system </span>

<span>y = -100 ft </span>
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3 years ago
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Can you help me with this??
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3 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

Distance covered in the slowing down phase = 250 m

v^2-u^2=2as

a= \frac {(v^2-u^2)}{2s}

a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2

v=u+at

t= \frac {(v-u)}{a}

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

The car is in the pit stop for 5s t_2=5 s

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

v=u+at

t= \frac{(v-u)}{a}

t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

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