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blondinia [14]
3 years ago
10

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, w

ith a standard deviation of σ = 0.5°C. How many readings must be taken so that the probability is 0.90 that the average is within ±0.1◦C of the actual temperature? Round the answer to the next largest whole number.
Physics
1 answer:
Viktor [21]3 years ago
3 0

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

0.9 = P(|\bar{x}-\mu|

0.9 = P(\frac{|\bar{x}-\mu|}{\frac{\sigma}{\sqrt{n}}}

0.9 = P(|Z|

For a confidence level of 90% our Z_{critic} is 1.645

Therefore,

\frac{0.1}{\frac{\sigma}{\sqrt{n}}} = 1.645

Substituting for \sigma = 5 and re-arrange for n, we have that n is equal to

n=(\frac{1.645\sigma}{0.1})^2

n=\frac{(1.645)^2(0.5)^2}{0.1^2}

n=67.65

n=68

We need to make 68 readings for have a probability of 90% and our average is within 0.1\°\frac

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C.) Monsoons.
8 0
3 years ago
Read 2 more answers
Two parallel, circular conducting plates 32 cm in diameter are separated by 0.5 cm and have charges of +24 nC and -24 nC, respec
inysia [295]

Answer: E = 33762.39 N/c

Explanation: we calculate the capacitance of the two conducting plates ( this is because, 2 circular disc carrying opposite charges and the same dimension form a capacitor).

C = A/4πkd

Where C = capacitance of capacitor

A = Area of plates = πr² ( where r is radius which is half of the diameter)

K = electric constant = 9×10^9

d = distance between plates = 0.5cm = 0.005 m

Let us get the area, A = πr², where r = D/2 where D = diameter

r = 32/2 = 16cm = 0.16m

A = 22/7 × (0.16)² = 0.0804 m²

By substituting this into the capacitance formula, we have that

C = 0.0804/4×3.142*9×10^9 × 0.005

C = 0.0804/565486677.646

C = 142.17*10^(-12) F.

But C =Q/V where V = Ed

Hence we have that

C = Q/Ed

Where C = capacitance of capacitor = 142.17*10^(-12)F

Q = magnitude of charge on the capacitor = 24×10^-9c

E = strength of electric field =?

d = distance between plates = 0.005m

142.17*10^(-12) = 24 ×10^-9 / E × 0.005

By cross multiplying

142.17*10^(-12) × E × 0.005 = 24 ×10^-9

E = 24 ×10^-9 / 142.17*10^(-12) × 0.005

E = 33762.39 N/c

8 0
3 years ago
The moon's gravity is one-sixth that of the earth. what is the period of a 2.00m long pendulum on the moon
erik [133]

Answer:

6.96 s

Explanation:

The period of a simple pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum and g the acceleration due to gravity.

In this problem, we have a pendulum with length L = 2.00 m, while the acceleration due to gravity is 1/6 that of the earth:

g' = \frac{g}{6}=\frac{9.8 m/s^2}{6}=1.63 m/s^2

So, the period of the pendulum on the moon is

T=2 \pi \sqrt{\frac{2.00 m}{1.63 m/s^2}}=6.96 s

3 0
3 years ago
A 76 N crate is hung from a spring
lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0
3 years ago
The potential energy of an object attached to a spring whose elastic constant is k is given by the formula p e space equals spac
Mademuasel [1]
You know the following:
k= 25N/m
x=1.5m
Now you just plug these numbers into the formula which is
Epe = 1/2kx²
       = 1/2(25N/m)(1.5m)²
       =  28.125J or in proper sigdigs 28J 
Hope this helps!
5 0
4 years ago
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