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marishachu [46]
3 years ago
12

PLEASE ANSWER You are in a vehicle waiting at a railroad crossing and notice a train going by at a constant speed of 5 kph. You

happen to notice that there’s no engineer driving the train and it seems to be on auto pilot. Down the track in the direction the train is traveling, you see a Humane Society van full of cute puppies and kittens stalled out on the tracks. Are the van’s passengers in trouble or not? Use Newton’s 2nd Law to explain.
Physics
2 answers:
Dominik [7]3 years ago
5 0

Reeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Kazeer [188]3 years ago
3 0
Newton’s second law is which simply states that the rate of change of momentum of a body over time is directly proportional to the force applied. An example is baseball. If you hit the ball with small force it won’t go far or fly in the air fast rather slowly glide through the air and probably touch the ground in 5 seconds. Based on that this question isn’t able to use the 2nd law. A train with no driver to stop it that weighs tons more than a van with puppies in it will destroy and plow through the van no matter what. It doesn’t matter the speed nor how far away it is. They are in danger due to the fact there is no driver to stop the train and there is a big mass difference between the two transportation devices.
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What would be new resistance if length of conductor is doubled and thickness is halved <br>​
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Answer:

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In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the
sp2606 [1]
  <span>I'll tell you how to do it but you must crunch the numbers. 

Use Kepler's 3rd Law 

T^2 = k R^3 

where k = 4(pi)^2/ GM 
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pi = 3.14159265 
T =3.09 days = 266976 seconds 
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a) 
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b) 
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6 0
3 years ago
how much energy is required to move a 4.00-microcoulomb charge through a potential difference of 36.0 volts?
musickatia [10]

Answer:

All you gotta do is search it up. ask the same question you asked on here, type it into google, it'll teach you a lot more than your actual teachers.

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4 0
3 years ago
What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
PolarNik [594]

Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

Also, \Delta E=h\times \nu

So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

8 0
3 years ago
Line d represents water. if the atmospheric pressure in a flask is lowered to 70 kpa, water would boil at what temperature?
sp2606 [1]
The diagram is in the picture attached.
Options are:
A) 32 °C
B) 70 °C
C) 92 °C
D) 100 °C

In order to find the value required, you need to look at the diagram and follow these steps:
1) search for the value of 70 kPa on the y-axis;
2) move on a horizontal line towards the right until you reach the line D;
3) move on a vertical line down, towards the x-axis;
4) read at what value of °C you are at.

Doing so, you can see that you are at a value a little bit above 90 °C (see picture).

Hence, the correct answer is C) 92°C.

5 0
3 years ago
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