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vampirchik [111]
3 years ago
10

Which of the following describes the magnetic field produced by a current carrying wire? Assume the wire is normal to the page,

and that current is flowing into the page.
A.) The magnetic field is directly into the page.

B.) The magnetic field is directly out of the page.

C.) The magnetic field surrounds the wire like a tube, with a clockwise field direction.

D.) The magnetic field surrounds the wire like a tube, with a counterclockwise field direction.
Physics
1 answer:
Maslowich3 years ago
6 0
D) the magnetic field surrounds the wire like a tube , with a counterclockwise field direction...
as it is in the left hand thumb rule
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HELP ASAP PLZZZZZZ
malfutka [58]
B because the electrons are moving to create a new form. Chemical energy is changing the form.


Hope this helps.
5 0
4 years ago
The frequency of the middle c note on a piano is 261.63 hz. What is the wavelength of this note in centimeters? The speed of sou
Sholpan [36]

Answer:

1.31 m

Explanation:

The relationship between frequency and wavelength of a sound wave is

c=f \lambda

where

c is the speed of the wave

f is the frequency

\lambda is the wavelenfth

In this problem, we have

c = 343.06 m/s

f = 261.63 Hz

So we can solve the formula for the wavelength:

\lambda=\frac{c}{f}=\frac{343.06 m/s}{261.63 Hz}=1.31 m

8 0
3 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
4 years ago
How much energy is stored in a spring with a spring constant of 390 N/m if the spring is compressed a distance of 0.45 m from it
masya89 [10]
F=-ks
F=-(390)(.45)
F=-175.5 N

Work=force x displacement
Work= 175.5(0.45)
Work= 78.98 J

Work = ∆E =78.98 J

Answer=79J (first option)
4 0
3 years ago
In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot
Fittoniya [83]

Answer:

Q=7.9\times 10^{-10}\ C

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

A=5\times 10^{-3}\ m^2

We Capacitance given as

For air

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}

C_1=2.2\times 10^{-11}\ F

C_2=\dfrac{K\varepsilon _oA}{d}

C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F

C_2=6.6\times 10^{-11}\ F

Net capacitance

C=C₁+C₂

C=8.8\times 10^{-11}\ F

We know that charge Q given as

Q= C V

Q=12\times 6.6\times 10^{-11}\ C

Q=7.9\times 10^{-10}\ C

6 0
3 years ago
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