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4 years ago

What would happen if you held the South Pole of one magnet near the North Pole of another magnet of the same size?

Physics

2 answers:

allsm [11]4 years ago

6 0

Answer:

(B) The magnets would be attracted to each other.

Explanation:

KiRa [710]4 years ago

4 0

The correct choice would be

(B) The magnets would be attracted to each other.

we know that like poles repel each other and unlike poles attract each other. hence when south pole of one magnet is placed near the north pole of other magnet, the two magnets attract each other due to interaction of magnetic fields of the two magnets. the magnetic field lines originating from north pole of one magnet ends at the south of other magnet

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A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second.

Leviafan [203]

Assuming Earth's gravity, the formula for the flight of the particle is:

s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. 

This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. 

Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet.

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4 years ago

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The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat

Nonamiya [84]

Answer:

E=0

Explanation:

The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero

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3 years ago

What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh

Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled s = 9.5 m.

Substituting these values in the equation

$s= u \timest+0.5at^{2}$

$t= \sqrt{\frac{2s}{g} }$

$t=\sqrt{\frac{2\times9.5}{9.8} }$

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

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3 years ago

Using planck's constant and a wavelength of 656e-9 what is the energy of the photon

lbvjy [14]

Answer:

3.03e-19 J

Explanation:

Use the formula E = hc/λ

Where:

h (Planck's constant) = 6.626e-34 J*s

c (speed of light, constant) = 3.00e8 m/s

λ (wavelength) = 656e-9 m

E = energy (in Joules)

E = (6.626e-34 * 3.00e8) / 656e-9 = 3.03018293e-19 = 3.03e-19 J

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3 years ago

The diagram above depicts an energy releasing process involving the uranium atom. Which form of energy does this process depict

ziro4ka [17]

I need help with this one to

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3 years ago

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