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mylen [45]
3 years ago
15

When a wire is attached to the negative terminal of a battery, what happens?

Physics
2 answers:
Artist 52 [7]3 years ago
5 0
Answer:





Step by step explanation
Leokris [45]3 years ago
3 0

Answer: The current will flow from the negative to the positive terminal as fast as possible. This will wear out the battery quickly.

Explanation:

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A ray of monochromatic light ( f = 5.09 × 10^14 Hz) passes from air into Lucite at an angle
harina [27]
     Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:

\frac{sen\O_{2}}{sen\O_{1}} = \frac{n_{2}}{n_{1}} \\ sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}}
 
     Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:

sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}} \\ sen\O_{2}= \frac{1.5* \frac{1}{2} }{1} \\ sen\O_{2}=0.75
  
     Using the arcsin properties, we get:

sen\O_{2}=0.75 \\ arcsin(0.75)=\O_{2} \\ \boxed {\O_{2}=48.59^o}

Obs: Approximate results, and the drawing is attached

If you notice any mistake in my english, let me know, because i am not native.

6 0
3 years ago
PlS help <br>I really need help.<br>very urgent<br><br>PLS help
horsena [70]
Draw a freebody diagram, it will explain it really well

the boat is floating on top of the water, which means that the net acceleration in the y direction must be zero

the boat is not sinking (dominant downwards acceleration/force)
the boat is not flying (dominant upward acceleration/force)

that measn
F_{net_y} =0

now, if you drew the FBD, you only have 2 forces acting on the boat.
the upward bouyancy force on the boat and the downward force due to weight

F_{net_y}= F_{bouyancy} - F_{weight}

since the net force is equal to zero

F_{bouyancy} - F_{weight} = 0

and thus

F_{bouyancy} = F_{weight}
4 0
3 years ago
Two identical balls each have a mass of 35.0 grams and a charge of . The balls are released from rest when they are separated by
OlgaM077 [116]

The given question is incomplete. The complete question is as follows.

Two identical balls each have a mass of 35.0 grams and a charge of q = 3.50 \times 10^{-6}C[/tex]. The balls are released from rest when they are separated by a distance of 6.00 cm. What is the speed of each ball when the distance between them has tripled? Use k = 9.00 \times 10^{9} Nm^{2}/C^{2}.

Explanation:

According to the conservation of energy, the formula will be as follows.

\frac{kq_{1}q_{2}}{r_{1}} = \frac{kq_{1}q_{2}}{(3r_{1})} + \frac{1}{2}mv^{2} + \frac{1}{2}mv^{2}

or,    \frac{kq_{1}q_{2}}{r_{1}}[1 - \frac{1}{3}] = mv^{2}

Putting the given values into the above formula as follows.

      \frac{kq_{1}q_{2}}{r_{1}}[1 - \frac{1}{3}] = mv^{2}

     \frac{9 \times 10^{9} \times (3.5 \times 10^{-6})^{2}}{0.09} \times \frac{2}{3} = \frac{35}{1000}v^{2}

           v^{2} = 23.333

                v = 4.83 m/s

Thus, we can conclude that speed of each ball when the distance between them has tripled is 4.83 m/s.

7 0
3 years ago
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