Please help this question.
1 answer:
Answer:
a). 59.049°C
b). 2.1179 seconds
Step-by-step explanation:
Expression representing the final temperature after decrease in temperature of the metal from 100°C to T°C is,
T =
where x = duration of cooling
a). Temperature when x = 5 seconds
T = 100(0.9)⁵
= 59.049
≈ 59.049°C
b). If the temperature of the metal decreases from 100°C to 80°C
Which means for T = 80°C we have to calculate the duration of cooling 'x' seconds
80 =
0.8 =
By taking log on both the sides
log(0.8) =log[]
-0.09691 = x[log(0.9)]
-0.09691 = -0.045757x
x =
x = 2.1179
x ≈ 2.1179 seconds
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Answer:
refer to the attachment please
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The Bruce must include X and Z in his report.
We have given the data on four electric circuits listed in the table
For w :
Voltage, V=18\ V
resistance,
The formula of Ohm's low is V=IR
I=0.47
So charge,q=It
q=3.76\ C
For x :
Voltage, V=24\ V
so, V=IR...(Ohm's low)
I=0.705
So charge,q=It
q=5.64\ C
For y :
Voltage, V=34\ V
resistance,
so, V=IR.......(Ohm's low)
I=0.485
So charge,q=It
q=3.88\ C
For z :
Voltage, V=12\ V
resistance,
so, V=IR......(Ohm's low)
I=0.66
So charge,q=It
q=5.28\ C
Therefore, the circuits X and Z deliver 5.64 C and 5.28 C of charge.
Hence, Bruce must include X and Z in his report.
To learn more about the electric circuit visit: