The patterns that can be looked for when studying science topic include the use of data or carrying out an experiment.
<h3>How to illustrate the information?</h3>
It should be noted that patterns are regular and intelligible forms or sequences that can be found in nature. Based on the information, scientific questions may be generated when scientists want to observe a pattern of events.
It should be noted that when scientists want to observe a pattern of events or
when something does not match an established pattern, scientists can use patterns to classify objects or phenomena into groups.
Therefore, the patterns that can be looked for when studying science topic include the use of data or carrying out an experiment.
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Answer:
The magnitude of the static frictional force is 1200 N
Explanation:
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x
= τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N
To solve this exercise it is necessary to take into account the concepts related to thermal expansion.
The thermal expansion is given by the function,

Where,
Change in Length
Change in Temperature
Coeficiente de dilatación lineal
Initial Length
By quickly deducing the formula, we can realize that the greater the change in temperature, the greater the change in the length of the radius.
The change in length is proportional to the change in temperature. Considering that the other two terms are constant we have that the correct one would be: <em>The hole in the center of the washer will expand.</em>
To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.
Let's start first by relating the amount of dose needed to be supplied per hour, in other words,
The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.




Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore



The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,


Therefore the client will receive 3.5mcg/kg/min.