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Ostrovityanka [42]
2 years ago
8

How would a small bar magnet be oriented when placed at position X?

Physics
1 answer:
Illusion [34]2 years ago
7 0

Based on the document attached, the  small bar magnet will be oriented when placed at position X is option d:

<h3>What is the magnet position about?</h3>

The orientation of the magnets   is known to be one that do affect its magnetic force. Atoms atoms, such as iron, are known to often gives a stronger magnetic force when compared to other atoms.

These atoms can be seen as tiny magnets, that has the north and south of each pole. If the poles is known to be oriented in all directions, the material will not bring about a net magnetic force.

Therefore, Based on the document attached, the  small bar magnet will be oriented when placed at position X is option d because the shapes of the magnetic field lines in regards to the bar magnet are said to be closed.

Learn more about magnet  from

brainly.com/question/12839896

#SPJ1

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A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land
erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + \frac{1}{2}at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = \sqrt{\frac{2h}{g} }.

⇒ t = \sqrt{\frac{2(1.3)}{9.81} } = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = \frac{s}{t}.

so, u = \frac{0.75}{0.51} = 1.45 m/s .

6 0
3 years ago
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

  max height is at  t= - b/2a = 10/9.8 =1.02

     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will <u>also</u> still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

8 0
2 years ago
Transform active margins are associated with which type of boundary? Convergent Transform Divergent
SSSSS [86.1K]

Answer:

Transform active margins are associated with which type of boundary?

Transform boundary

Explanation:

The transform boundary is a boundary where one plates(crust) slides past another plate horizontally. This kind of plate movement have been detected to exist between the interaction of the North pacific plates(continental plate) and the pacific plates(oceanic plates) .

At the transform margin the crust are usually broken. But overall crust are neither created nor destroyed . The transform margin region are active as it is marked by shallow-focus earthquakes .

Along the fractured zone where this transform movement occurs is known to create extensive transform faults .Notable transform fault that exist in this kind of boundary(transform) is the San Andrea fault and Alpine Fault.  

The motion of this plates can occur on a single fault or on a group of faults.

5 0
4 years ago
The maximum displacement of an oscillatory motion is A=0.49m. Determine the position x at which the kinetic energy of the partic
Tatiana [17]

Answer:

The position x, is ± 0.4 m.

Explanation:

The total mechanical energy of the oscillatory motion is given as;

E_T = U +K.E\\\\E_T = \frac{1}{2} kA^2\\\\ \frac{1}{2} kA^2 = U +K.E\\\\K.E = \frac{1}{2} kA^2 - U ----equation(1)

When the kinetic energy (E) is half of the elastic potential energy (U);

K.E = \frac{U}{2} ----equation(2)

Equate (1) and (2)

\frac{U}{2} = \frac{1}{2} kA^2 - U\\\\U = kA^2 -2U\\\\U+2U = kA^2\\\\3 U =  kA^2\\\\3(\frac{1}{2} kx^2) = kA^2\\\\\frac{3}{2} x^2=A^2\\\\x^2 = \frac{2}{3} A^2\\\\x = \sqrt{\frac{2}{3} A^2} \\\\x = A\sqrt{\frac{2}{3} } \\\\x = 0.49\sqrt{\frac{2}{3} }\\\\x = + /- (0.4 \ m)

Thus, the position x, is ± 0.4 m.

8 0
3 years ago
A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how
Leona [35]
<span>345/115=3 3*24=72 the answer is B. 72 </span>
8 0
3 years ago
Read 2 more answers
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