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kenny6666 [7]
3 years ago
11

Name three insulators

Physics
2 answers:
aliya0001 [1]3 years ago
7 0
Paper, rubber and glass are some examples of insulators.
Firlakuza [10]3 years ago
3 0
Bubble wrap, wood, plastic and rubber
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Answer:

The answer is 5 meters per second.

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3 years ago
Please help me on question 3a and 3b.<br><br>Thanks! ​
Sonbull [250]

(a) The frequency of water wave is 2 Hz.

(b) The wave speed of the water wave is 3.6 m/s.

<u>Explanation:</u>

(a) It is known that completion of one complete wave in 1 second is defined as frequency of 1 HZ. So here there are 120 waves crossing the boat in 1 minute. So the frequency of the water wave will be

            Frequency =\frac{\text { Number of waves }}{\text { Time in seconds }}

As the time is 1 minute which is equal to 60 seconds and the number of waves is given as 120 then the frequency of the water wave is

         \text { Frequency }=\frac{120}{60}=2 \mathrm{Hz}

So the frequency of water wave is 2 Hz.

(b) Then if the wavelength of the water wave is 1.8 m with a frequency of 2 Hz, then speed of the wave can be determined as the product of wavelength with frequency.

So Speed = Frequency × Wavelength

Speed = 2 × 1.8 = 3.6 m/s.

So the speed of the water wave is 3.6 m/s.

7 0
3 years ago
Insert
nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

A\propto N

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

A\propto m

where m is the mass of the sample.

In this problem:

- When the mass is m_1 = 1 g, the activity is A_1=16 count/min

- When the mass is m_2=20 g, the activity is A_2

So we can find A2 by using the rule of three:

\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min

B)

The equation describing the activity of a radioactive sample as a function of time is:

A(t)= A_0 e^{-\lambda t} (1)

where

A_0 is the initial activity at time t = 0

t is the time

\lambda is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

\lambda = \frac{ln2}{t_{1/2}}

where t_{1/2} is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

t_{1/2}=5700 y

So its decay constant is

\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

A_0 = 320 count/min (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min

C)

We know that a sample of living wood has an activity of

A=16 count/min per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

A_0 = A\cdot 5 = (16)(5)=80 count/min

Moreover, here we have a sample of 5 g, with current activity of A=20 count/min: it means that its activity per gram of mass is

A'=\frac{20}{5}=4 count/min

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

t=2t_{1/2}=2(5700)=11,400 y

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Is god real??? i wanna know
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