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a_sh-v [17]
3 years ago
13

What happens to the open circuit if a small fan is connected at point f and the circuit is closed

Physics
1 answer:
liraira [26]3 years ago
4 0
There's no way to tell.  Without seeing a diagram of the circuit,
I'll need to know much more about it than you've told me.
I don't know anything about the components or power supply
that are in the circuit, and I don't know where point ' f ' is in it.

Right now, even with the copious volume of all the available
information, no answer to your question is possible. 
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A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will
alukav5142 [94]

Answer:

The change  in momentum is  \Delta p =   kg \cdot m/s      

Explanation:

From the question we are told that  

       The mass of the probe is  m = 170 kg

       The location of the prob at time t = 22.9 s is  A  =

       The  momentum at time  t = 22.9 s is  p = < 51000, -7000, 0> kg m/s

        The net force on the probe is  F =  N

Generally the change in momentum is mathematically represented as

              \Delta p = F * \Delta t

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

           \Delta p =  * (22.9 - 22.6)

            \Delta p =  * (0.3)  

              \Delta p =   kg \cdot m/s        

 

6 0
3 years ago
Which of these is an experiment generally regarded as being first carried out by James Joule?
Sholpan [36]
1. The correct answer among the choices provided is the third option. Measuring the temperature increase of water from doing work by stirring it is an experiment generally regarded as being first carried out by James Joule.
2.  Joule's experiment directly shows that heat is a form of energy. He wanted to make a different way of measuring energy.
8 0
3 years ago
Read 2 more answers
18. When is an atom unlikely to react and form chemical bonds?
katrin2010 [14]
Fission and fusion are two processes that release very little amounts of energy.
5 0
3 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
gregori [183]

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 * 10^9years.

Answer:

The potassium-40 present in 80 kg is  Z = 0.0288 *10^{-3}\ kg

The effective dose absorbed per year is  x = 2.06 *10^{-24} per year

Explanation:

From the question we are told that

      The mass of potassium in 1 kg of human body is m =  3g= \frac{3}{1000} =  3*10^{-3} \ kg

      The mass of the person is M = 80 \ kg

       The abundance of Potassium-39 is   93.26%

        The abundance of Potassium-40 is   0.012%

         The abundance of Potassium-41 is   6.78 %

         The energy absorbed is  E =  1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J

Now  1 kg of human body contains       3.0*10^{-3}\ kg of  Potassium

So      80 kg of human body contains      k kg of  Potassium

=>   k = \frac{ 80 * 3*10^{-3}}{1}

     k = 0.240\  kg

Now from the question potassium-40 is  0.012% of the total  potassium so

     Amount of potassium-40  present is mathematically represented as

            Z = \frac{0.012}{100}  * 0.240

            Z = 0.0288 *10^{-3}\ kg

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

           D =  \frac{E}{M}

Substituting values

          D =  \frac{1.7622*10^{-13}}{80}

            D =  2.2*10^{-15} J/kg

Converting to Sieverts

We have

           D_s = REB * D

           D_s = 1.2 * 2.2 *10^{-15}

           D_s =  2.64 *10^{-15}

So

     for half-life (1.28 *10^9 \ years)  the dose is  2.64 *10^{-15}

     Then for 1  year the dose would be  x

=>         x = \frac{2.64 *10^{-15}}{1.28 * 10^9}

             x = 2.06 *10^{-24} per year      

7 0
3 years ago
Which type of graph most clearly show how a part or percentage relate to the whole
Bezzdna [24]

Circle graph is the type of graph which most clearly shows the part or percentage related to the whole.

<u>Explanation:</u>

Graphs are the easiest tool to show the change in a given dependent variable with respect to independent variable. The influence or the effects will be seen in the graphs.

There are different kinds of graph like circle graph, bar graph, line graph. Each of these have their own method of showing the change in the dependent variable. Among them, circle graph is the one which will help to show the percentage of each variable related to the whole.

6 0
3 years ago
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