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MAVERICK [17]
2 years ago
9

When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in f

lux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?
Physics
1 answer:
Pepsi [2]2 years ago
7 0

Answer:

<em>The net force is directed downwards</em>.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

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Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?
natulia [17]
If a galaxy is located 200 million light years from Earth, you can conclude that t<span>he light will take 200 million years to reach Earth. </span>
8 0
2 years ago
Read 2 more answers
A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha
Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

PQ = \sqrt{4^{2}+3^{2}}=5 m

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

3 0
2 years ago
How can we use the balloon experiment to prove that air has weight (even though we cannot see air)?
Gala2k [10]

Answer:

The end of the meter stick with the deflated balloon should have risen into the air. ... The only way the balloon could have lost mass is if the air that was inside it has mass. With this experiment you have shown that air takes up space and has mass, so you have proven that air is matter.

Explanation:

7 0
2 years ago
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A 4.99 m long rod of negligible weight is attached on one end to a ball joint which allows the rod to rotate in all directions.
Aleksandr [31]

Answer:

The answer is 91.18 Nm

Explanation:

Solution

Recall that

The length of the rod = 4.99 m

∅ = 26°

Force = 62.5N

Now,

T = r * F

The direction of the torque will be in horizontally northward

The torque magnitude is  T =r F sin θ

where ∅ will be the angle between r + F  θ= 163°

Therefore,

T = 4.99 * 62.5 * sin 163

T =91.18 Nm

6 0
2 years ago
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