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malfutka [58]
3 years ago
10

Cuando es más confiable el valor central de una medición

Physics
1 answer:
Olin [163]3 years ago
5 0

Answer:

La media es la medida de tendencia central más utilizada y, en general, se considera la mejor medida de la misma. Sin embargo, hay algunas situaciones en las que se prefiere la mediana o la moda.

La mediana es la medida preferida de tendencia central cuando:

Hay algunas puntuaciones extremas en la distribución de los datos.

Explanation:

Espero que esto ayude a marcar el MÁS CEREBRAL !!!

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A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
Ratling [72]

Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

6 0
3 years ago
What do we call air in the motion​
r-ruslan [8.4K]
Answer: i think its wind
3 0
3 years ago
Read 2 more answers
A train is accelerating at a rate of 2 km/hr/s.  If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?
Mademuasel [1]
Here it is. *WARNING* VERY LONG ANSWER

________________________________________... 
<span>11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) </span>

<span>The change in PE =mgh=5*9.8*12=588 J </span>
<span>______________________________________... </span>
<span>12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. </span>

<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>

<span>Their KE as they crossed the line=(1/2)Mv^2 </span>

<span>Their KE as they crossed the line=0.5*550*(15.98)^2 </span>

<span>Their KE as they crossed the line is 70224.11 J </span>

<span>______________________________________... </span>
<span>13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m </span>

<span>She trips and drops the spare tire of mass = m = 10.0 kg, </span>

<span>The tire rolls down the hill with an intial speed = u = 2.00 m/s. </span>

<span>The height of top of the next hill = h = 5.00 m </span>

<span>Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 </span>

<span>Initial total mechanical energy =mgH+(1/2)mu^2 </span>

<span>Suppose the final speed at the top of second hill is v </span>

<span>Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 </span>

<span>As mechanical energy is conserved, </span>

<span>Final total mechanical energy =Initial total mechanical energy </span>

<span>mgh+(1/2)mv^2=mgH+(1/2)mu^2 </span>

<span>v = sq rt [u^2+2g(H-h)] </span>

<span>v = sq rt [4+2*9.8(20-5)] </span>

<span>v = sq rt 298 </span>

<span>v =17.2627 m/s </span>

<span>The speed of the tire at the top of the next hill is 17.2627 m/s </span>
<span>______________________________________... </span>
<span>14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. </span>

<span>a.)The mass of bean = m = 2.0 g </span>

<span>Height up to which the been jumps = h = 1.0 cm from hand </span>

<span>Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg </span>

<span>b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s </span>
<span>_____________________________ </span>
<span>15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. </span>

<span>The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s </span>
<span>__________________________________ </span>
<span>16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, </span>

<span>The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 </span>
<span>______________________________________... </span>

<span>EDIT </span>
<span>1.) A train is accelerating at a rate = a = 2.0 km/hr/s. </span>

<span>Acceleration </span>

<span>Initial velocity = u = 20 km/hr, </span>

<span>Velocity after 30 seconds = v = u + at </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr </span>

<span>Velocity after 30 seconds = v = 80 km/hr=22.22 m/s </span>
<span>_______________________________- </span>
<span>2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. </span>

<span>His acceleration = a =11.1/9=1.233 m/s^2 </span>

<span>Distance he covered = s = (1/2)at^2=49.95 m</span>
7 0
3 years ago
Two technicians are discussing a fuel gauge on a General Motors vehicle. Technician A says that if the ground wire's connection
r-ruslan [8.4K]

Answer:

b. Technician B only

Explanation:

There is a float connected to the variable resistor in a fuel tank unit. The resistance of the variable resistor also changes as the fuel level changes. The tank unit's resistance changes, the dash mount gauge also changes and available on driver's display.  

If the tank transmitter is disconnected, the operation will not take place and the resistance change will not be transmitted to the dash unit. The needle will therefore remain the empty one at all times.Even after being rusty, the ground wire connection to the fuel tank will be able to conduct. Hence Technician B   is correct and Technician A is incorrect.

8 0
3 years ago
It take my planet amount of time to orbit the sun
navik [9.2K]
It takes the Earth 365 days to orbit the sun.
4 0
3 years ago
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