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Fantom [35]
3 years ago
5

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 gram

s. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?
Physics
1 answer:
Bogdan [553]3 years ago
4 0

Complete Question

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?

A His results are accurate but not precise.

B His results are neither accurate nor precise.

C His results are both accurate and precise

D His results are precise but not accurate.

Answer:

Correct option is A

Explanation:

From the question we are told that  

   The  yield of the first test k  =  5.2 \  g

   The  yield of the second  is  u =  1.3 \  g

   The  third yield is  p =  8.5 \  g

   The  average yield  A = 5.0 \ g

    The  know yield is  A_S =  5.1 \  g

From the data given we see that

        A_S \ne A

Since his average yield is closer to the known yield then the answer is accurate

But since the yield for each test are not repeated the answer is not precise

So the answer is accurate but not precise  

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Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
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Answer:

a) (dT/dt) = -0.3 [T - 70]

b) (dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) (dT/dt) = -18 {T - [66 cos (2πt)]}

with t in hours

d) (dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Explanation:

The Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.

If the temperature of the object = T

Temperature of the surroundings = Ambient temperature = TA(t)

(dT/dt) ∝[T - TA(t)]

Introducing the constant of proportionality, k

(dT/dt) = k [T - TA(t)]

Temperature is in degree Celsius and time is in minutes.

Because the temperature of the body is decreasing, we introduce a minus sign

(dT/dt) = -k [T - TA(t)]

a) If TA(t) = 70°C, k = 0.3

(dT/dt) = -0.3 [T - 70]

b) The ambient temp TA(t) = 66 cos ((π/30)t) degrees Celsius (time measured in minutes).

(dT/dt) = -k [T - TA(t)]

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

(dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) If we measure time in hours the differential equation in part (b) changes.

1 hour = 60 mins

If t is now expressed in hours,

t hours = (60t) mins

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

dT = -k {T - [66 cos ((π/30)t)]} dt

dT = -k {T - [66 cos ((π/30)60t)]} d(60t)

(dT) = -60k {T - [66 cos ((π/30)60t)]} dt

(dT/dt) = -60k {T - [66 cos (2πt)]}

with t in hours, k = 0.3, 60k = 18

(dT/dt) = -18 {T - [66 cos (2πt)]}

d) If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more.

If T is in degree Fahrenheit

T°F = (5/9)(T°F - 32) degrees Celsius

T°F = [(5T/9) - 17.78] degrees Celsius

(dT/dt) = -60k {T - [66 cos (2πt)]}

time already converted to hours.

dT = -60k {T - [66 cos (2πt)]} dt

66 cos (2πt) degrees Celsius = {(9/5) [66 cos (2πt)] + 32} degrees Fahrenheit = {[118.8 cos (2πt)] + 57.6} degrees Fahrenheit

d[(5T/9) - 17.78] = -60k {T - [118.8 cos (2πt) + 57.6]} dt

(5/9) dT = -60k [T - 57.6 - 118.8 cos (2πt)] dt

(5/9) (dT/dt) = -60k [T - 57.6 - 118.8 cos (2πt)]

(dT/dt) = -108k [T - 57.6 - 118.8 cos (2πt)]

k = 0.3, 108k = 32.4

(dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Hope this Helps!!!

7 0
4 years ago
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