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alexandr402 [8]
3 years ago
8

Why is it important to consider experimental error in all the empirical results presented?

Physics
1 answer:
NemiM [27]3 years ago
3 0
Because when your running any experiment there will always be an experimental error so always be ready for it so that you can correct that error in your experiment
You might be interested in
Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8
sergejj [24]

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

    T_1 =  98 F

  T_2 =  39^oC

  T_3 =  303 \  K

Generally converting T_3 to  Fahrenheit

    T_3' =  (T_3 -273 ) * \frac{9}{5}  + 32

=> T_3' =  (303 -273 ) * \frac{9}{5}  + 32

=> T_3' = 86 F

Converting  T_2 to  Fahrenheit

      T_2' =  T_2 * \frac{9}{5}  + 32

=> T_2' =  39 * \frac{9}{5}  + 32

=> T_2' =102.2 F  

Now comparing  the temperature  in Fahrenheit we see that T_2  is the hottest

3 0
3 years ago
An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into
Aleksandr-060686 [28]

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

\rho = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N

The drag force on the athlete is 33 N

3 0
3 years ago
Round to the hundredths place.
Ivanshal [37]

Answer:

21 protons

Explanation:

4 0
3 years ago
328,000 mm + 137 m = ______________m?
svp [43]

Answer:

465m.

Explanation:

Convert all units to meters. So,

328 + 137 = 465m.

5 0
3 years ago
Please help me with this question guys.
katen-ka-za [31]

Answer:

<em>The average speed is 22.2 km/h</em>

Explanation:

<u>Average Speed</u>

Given an object travels a total distance d and took a total time t, then the average speed is:

\displaystyle \bar v=\frac{d}{t}

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h

Then he drives d2=7 km at v2=43 km/h taking a time of:

\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h

The average speed is 22.2 km/h

7 0
3 years ago
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