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lana66690 [7]
3 years ago
8

Short-range forecasts tends to ________ longer-range forecasts.

Physics
1 answer:
melisa1 [442]3 years ago
6 0
<span>Short-range forecasts are more accurate than longer range ones. Short term forecasts may use mathematical techniques such as moving averages and, exponential smoothing. Longer term forecasts not only use different methodologies, such as qualitative vs. quantitative, they also tend to consider different issues.</span>
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Why is heat acclimatization important?
kicyunya [14]

Answer:

Heat acclimatization :

   It is the biological adaptations or we can say that it coverts according to the present environment.It also reduce the strain and maintain the normal temperature and heart rate.Heat acclimatization also increase the comfort and reduce all the mental strain and also protect out liver ,muscles ,kidneys and brain fro the injury.

5 0
3 years ago
Power is measured in Joules per second, which is also the<br> _____
Monica [59]

Answer:

Watt

Explanation:

Power is measured in Watts. J/s is the base unit of measurement, but we usually measure power in Watts (W).

3 0
3 years ago
A car travels for 5.5 h at an average speed<br> of 75 km/h. How far did it travel?<br> T TE
ANEK [815]

Answer:

\boxed {\boxed {\sf 412.5 \ kilometers }}

Explanation:

Distance is the product of speed and time.

d=s*t

The speed of the car is 75 kilometers per hour. It traveled for 5.5 hours.

s= 75 \ km/hr \\t= 5.5 \ hr

Substitute the values into the formula.

d= 75 \ km/hr * 5.5 \ hr

Multiply. Note that the hours will cancel each other out.

d= 75 km * 5.5 \\d= 412.5 \ km

The car travelled <u>412.5 kilometers.</u>

5 0
2 years ago
An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then
gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

a=\frac{dv}{dt}=1.76

So, the velocity can be obtained by integrating this expression:

v=1.76t

The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

At this time the velocity is: v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}

This velocity remains constant in the section 2, so for that section the movement equation is:

x=v*t\\t=\frac{x}{v}

The left distance is 89 meters, and the velocity is 6.2225\frac{m}{s}, so:

t=\frac{89}{6.2225}=14.303s

So, the total time is 14.303+3.5355s=17.838s

7 0
2 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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