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dexar [7]
4 years ago
13

You should increase the distance between your car and the vehicle ahead when you:

Physics
1 answer:
ELEN [110]4 years ago
8 0
<span>You should increase the distance between your car and the vehicle ahead when you are being tailgated by another driver.
Tailgated is a condition when the vehicle behind you is very close to you that there is no distance between your vehicle and the vehicle behind you. In that case you should increase distance from the vehicle ahead to avoid collision in any case. These days people were mostly in hurry and everyone want to go ahead, so tailgating is very common.</span>
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The process of recording the strength of the contraction of a muscle when it is stimulated by an electrical current is known
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i dont know the answer of your question correctly

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4 years ago
Why is alternating current more effective at long–distance travel than direct current?
amid [387]
Because we can send high voltage, low current which is more efficient since large amounts of current in the line causes power inefficiencies. AC can be transformed locally to high current lower voltage
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3 years ago
Read 2 more answers
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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In which direction are the winds blowing in the northern and southern hemispheres
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The winds normally blow from east to the west rather than blowing from the north to the south in the northern and the southern hemispheres.

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The reason for the blowing of the wind from the east to the west rather than from the north to the south in both the hemispheres which are the northern and the southern hemispheres are because it happens in light of the fact that Earth's rotation around it's own axis creates what is known as the Coriolis effect.

The Coriolis impact makes wind frameworks turn counter-clockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere.

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