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Sladkaya [172]
3 years ago
12

a bucket of water of mass 20kg is pulled at a constant velocity up to a platform 40 meters above the ground. this takes 10 minut

es , during which time 5kg of water drips out at a steady rate through a hole in the bottom . find the work required to raise the bucket to the platform.
Physics
1 answer:
erma4kov [3.2K]3 years ago
5 0
<span>Taking into account the information above, we know the average mass of the bucket of water may be m=20-5/2=17.5kg. As the bucket of water is pulled at a "constant velocity" the work required to raise the bucket to the platform transformed into the potential energy of the bucket of water. That is why it should be W=mgh=17.5*9.8*40=6860J</span>
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Acceleration problem <br> Show work plz
Dennis_Churaev [7]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)  

vo² = 630               simplifying

vo = 25 m/s            result

3 0
3 years ago
In addition to average weather conditions, climatological data also describes annual variations and fluctuations of temperature,
drek231 [11]

<u>Answer:</u>

In addition to average weather conditions, climatological data also describes annual variations and fluctuations of temperature, precipitation, wind speed and other variables.

<u>Explanation</u>:

A lot many observations are made all around the world regarding the weather each day. These observations and analysis are done by humans as well as automated instruments. The weather data is collected each day all year and any inaccuracies and discrepancies are checked and rectified. The results are later then presented as the climate data. There are various factors that are taken into consideration while determining the climate of a region. Apart from the factors that are already mentioned, wind speed is also one of the other variables.

8 0
3 years ago
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suter [353]
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The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr
Dvinal [7]

Answer:

T_A_v_g=9.918192559$^{\circ}C

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

T(t)=10-5*sin(\frac{\pi}{12t})

The average temperature over a given interval can be calculated as:

T_a_v_g=\frac{T_o+T_f}{2}

Where:

T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C

T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C

Hence, the average temperature between noon and midnight is:

T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C

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3 years ago
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Answer:

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Explanation:

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