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Aliun [14]
3 years ago
9

A bar magnet is attached solidly to a frictionless surface and its length is aligned with the x axis.To the right of the first m

agnet a short distance away is a second bar magnet with its center placed on the x axis and its length perpendicular to the x axis.The second magnet is free to move.Once placed in position at rest,which best describes the initial motion of the second magnet?
a. The magnet will move away from the fixed magnet.
b. The magnet will not move.
c. The magnet will move toward the fixed magnet.
d. The magnet will start to rotate.
Physics
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

the correct answer is D

Explanation:

For this exercise we must remember that the poles of the same if not repel each other and the poles of different signs attract.

With this, let's analyze the situation presented.

The two magnets are perpendicular, with the second magnet to the right of the first.

We have two cases:

* first magnet with the north pole to the right

  If the north of the second magnet is upwards there is a repulsion and the south pole of the second magnet there is an attraction with the north pole of the first magnet, so it would have a force that has to rotate the second magnet.

The force with the south pole of the first magnet that is at a greater distance is less, so the resultant of force is determined by the nearest poles.

If the poles of the second magnet are reversed, that is, the South pole up and the North pole down, the same result is obtained, but with a twist in the opposite direction.

* first magnet with the south pole to the right

    Repels with the south pole of the second magnet and attracts with the north of the second magnet

therefore in both possibilities the second magnet acquires a rotational movement

Consequently the correct answer is D

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A clarinet sounds as a closed pipe. if a clarinet sounds a note with a pitch of 375 hz, what are the frequencies of the lowest t
mote1985 [20]
So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.

Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.

So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
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etc (: I hope this helps.
8 0
4 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
A(n) 1946 kg car travels at a speed of 10 m/s . What is its kinetic energy ? Answer in units of J.
erica [24]

Answer:

KE=97300J

Explanation:

KE=1/2mv^2

KE=1/2(1946)(10)^2

KE=97300J

4 0
3 years ago
Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz and face each other at a distance of 1.24 m. Loc
Juli2301 [7.4K]

Answer:

0.2286  m,  0.686 m and  1,143 m

therefore we see that there is respect even where the intensity is minimal

Explanation:

Destructive interference to the two speakers is described by the expression

       Δr = (2n +1) λ/2

where r is the distance,  λ the wavelength and n an integer indicating the order of the interference

let's locate the origin on the left speaker

let's find the wavelength with the equation

        v =  λ f

         λ = v / f

we substitute

      Δr = (2n + 1) v / 2f

let's calculate for difference values ​​of n

     Δr = (2n +1) 343/(2 750)

     Δr = (2n + 1) 0.2286

we locate the different values ​​for a minimum of interim

    n     Δr (m)

     0    0.2286

     1     0.686

     2    1,143

therefore we see that there is respect even where the intensity is minimal

6 0
3 years ago
A worker wants to load some bricks in to his van. there are 1000 bricks and when stacked neatly they measure 2m by 1m by 1m, wha
inessss [21]

lenddbisnwjw whwjwiwnww

8 0
3 years ago
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