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nadezda [96]
3 years ago
10

A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive dir

ection. What is the magnitude of the minimum magnetic field needed to levitate the rod
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

Magnetic field, B = 0.23 T          

Explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.

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A constant net torque is exerted on an object. Which of the following quantities for the object cannot be constant? (Select all
prohojiy [21]

Answer:

A. kinetic energy

B. angular velocity

E. angular position

Explanation:

The quantities that cannot be constant if a constant net torque is exerted on an objecta are:

A. Kinetic energy. If a torque is applied, the linear or angular speed will be changing at a rate proportional to the torque, so the kinetic energy will change too.

B. Angular velocity. It will change at a rate equal to the torque.

C. Angular position. If the angular velocity changes, the angular position will change.

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What happens when two minerals have different arrangements of Atoms
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The overall arrangements of the atoms produce crystals
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Fan blades would be an analogy for which atomic model? Thomson's Bohr's electron cloud Dalton's
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The correct answer would be Electron Cloud 
5 0
2 years ago
Please help . Anybody
Alona [7]

Answer:

1: c

2: a

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6 0
2 years ago
A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. How far did the ball travel hor
Genrish500 [490]

Answer:

414.9 m

Explanation:

First, become familiar with the horizontal, and vertical vector components.

Vertical vector: Vy = V × sin (θ).

Horizontal vector: Vx = V × cos(θ).

Distance traveled = Velocity vector × time in the air.

Time in the air given Vy = 2 × Vy / g (in respect to the metric of the vector).

Range of the projectile = Vx² / g

Time in the air given Vx = (Vx + √(Vx)² + 2gh) / g.

Given a 28° angle with an initial velocity of 70m/s, we have enough information to calculate!

Vx = 70 m/s × cos(28°) ≈ 61.806 m/s

Vy = 70 m/s × sin(28°) ≈ 32.863 m/s

t = 2 × Vy / g

t = 2 × ≈32.863 / 9.8

t = ≈65.726 / 9.8

t ≈ 6.7 s

Distance traveled (horizontal) = Vx × t = 61.806 × 6.7 ≈ 414.9 m

6 0
2 years ago
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