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andre [41]
3 years ago
11

The magnetic field or force seems to be associated with the lineup of____________ within the magnet.

Physics
2 answers:
Nastasia [14]3 years ago
6 0
The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
Luda [366]3 years ago
6 0

Answer:

non of the answers listed are right. the correct answer is domains

Explanation:

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Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car i
Korvikt [17]

As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

\Delta x . \Delta P = \frac{h}{4\pi}

\Delta x . m \Delta v = \frac{h}{4\pi}

now plug in all data

(5\times 0.3048). (2250 \times 0.454) \Delta v = \frac{6.6 \times 10^{-34}}{4\pi}

\Delta v = 3.37 \times 10^{-38} m/s

So above is the uncertainty in velocity of the object

4 0
3 years ago
Read 2 more answers
Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the
sdas [7]

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

         F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

6 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f
Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

  • According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
  • Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
  • Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

       W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)

  • As we have already said, (1) is equal to the final kinetic energy of the puck:
  • ⇒ Kf = 470 mJ  (2)
8 0
3 years ago
An object with a lot of mass has ____________.
aleksley [76]
C. a lot of inertia.............................. :-)
3 0
3 years ago
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