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ioda
1 year ago
11

make a sketch showing the direction of the magnetic field from the solenoid. on your sketch, label the induced north magnetic po

le and induced south magnetic pole in the iron
Physics
1 answer:
leonid [27]1 year ago
8 0

A coil of wire with electric current running through it is a solenoid. A solenoid is depicted in the Figure below.

A magnetic field with north and south poles, similar to a bar magnet, is created as current flows through the coil. The magnetic field inside the solenoid has a parallel, linear pattern. A solenoid's magnetic field moves from its north pole to its south pole in one direction only. The north and south poles of a solenoid can be determined using the clock face rule. A solenoid is enclosed by a magnetic field and has north and south magnetic poles similar to a bar magnet.

To learn more about solenoid

brainly.com/question/15504705

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Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?
nikitadnepr [17]

Answer:

The appropriate response is "Optical printer ".

Explanation:

  • A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses.
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4 0
4 years ago
Volcanoes tend to erupt at places where
tigry1 [53]

Answer:

it is 3

Explanation:

because the crack will be open for the magma to come out

4 0
3 years ago
Read 2 more answers
10. A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant
vladimir1956 [14]
The velocity of the ball is 30.0 m/s.
5 0
3 years ago
A spring has a natural length of 28.0 cm. If a 23.0-N force is required to keep it stretched to a length of 36.0 cm, how much wo
krek1111 [17]

Answer:

0.23 J

Explanation:

k*(36 - 28) = 23

so k = 23/8 N/cm

W = k(32 - 28)²/2 = 23/8 * 4²/2 = 23 N-cm = 0.23 J

4 0
2 years ago
A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
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