Answer:
Explanation:
Given
n=5
0.3 fraction recrystallize after 100 min
According to Avrami equation

where y=fraction Transformed
k=constant
t=time


Taking log both sides


At this Point we want to compute 



taking log both sides



Rate of Re crystallization at this temperature

Yes heating water allows it to dissolve more Sugars because the molecular distance increases and this distance can be covered by more sugar. In the given question, The independent variable would be the temperature of water.
Since to whatever temperature the water boils at the boiling temperature of does not change remains hundred degree. Rest all the variables can vary the weight of the amount of sugar with the variable in the temperature of Boiling of water to remain constant.
The starter motor's potential difference across the headlight bulbs is 38.45V, requiring an additional 39 a from the battery. Voltage, also known as potential difference.
It is sometimes described as the amount of work needed to move a test charge between two sites, expressed as a unit of charge. Volt is the potential difference's SI unit (V). We only take into account the charge between the locations P and Q when current moves between them in an electric circuit. Electric potential difference between two sites is referred to as voltage, also known as electric pressure, electric tension, or (electric) potential difference. an electric field that is static.
Vh = I*Rn
Vh = 39/5.476*5.40v
Vh = 38.45v
Learn more about voltage here
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To solve for distance use the formula for distance d = st, or distance equals speed times time.
distance = speed x timeSpeed
so the answer is 240 miles per hour.
Answer:
Explanation:
A.
Given:
Vo = 21 m/s
Vf = 0 m/s
Using equation of Motion,
Vf^2 = Vo^2 - 2aS
S = (21^2)/2 × 9.8
= 22.5 m.
B.
Given:
S = 22.5 + 21 mm
= 22.521 m
Vo = 0 m/s
Using the equation of motion,
S = Vo × t + 1/2 × a × t^2
22.521 = 0 + 1/2 × 9.8 × t^2
t^2 = (2 × 22.521)/9.8
= 4.6
t = 2.14 s