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koban [17]
3 years ago
9

Assume that a particle moves in a velocity field defined by V (x, y) = (x + 6)i + (y − 3)j. Find the path of a particle with ini

tial starting point (1, 1).
Physics
1 answer:
gladu [14]3 years ago
8 0

Answer:

V(1,1)=7i-2j

Explanation:

V(x,y)=(x+6)i+(y-3)j ............eq1

As intial point are given (1,1)

which means x=1,y=1

put these values in eq1

V(1,1)=(1+6)i+(1-3)j

V(1,1)=7i-2j

And its magnitude given as

|V(1,1)|=\sqrt{(7)^{2}+(-2)^{2}  }

|V(1,1)| =7.28

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3 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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3 years ago
A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef
AveGali [126]
The thermal efficiency of an engine is
\eta= \frac{W}{Q}
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W is the work done by the engine
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