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Pani-rosa [81]
3 years ago
11

Trina makes a diagram to summarize how to use the right-hand rule to determine the direction of the magnetic force on each type

of charge.
Which labels belong in the regions marked X, Y, and Z?

A) X: Thumb, for velocity
Y: Fingers, for magnetic field
Z: Palm of hand, for force

B) X: Back of hand, for force
Y: Thumb, for velocity
Z: Palm of hand, for force

C)X: Palm of hand, for force
Y: Thumb, for velocity
Z: Fingers, for magnetic field

D) X: Palm of hand, for force
Y: Fingers, for magnetic field
Z: Back of hand, for force

Physics
2 answers:
Tems11 [23]3 years ago
7 0

Labels that belong in regions X, Y, Z;

B) X: Back of hand, for force

Y: Thumb, for velocity

Z: Palm of hand, for force

mel-nik [20]3 years ago
6 0

Answer: A)

B) X: Back of hand, for force

Y: Thumb, for velocity

Z: Palm of hand, for force

Explanation:

Right-hand rule is used to find direction of magnetic force. When a charge enters magnetic field, magnetic force acts perpendicular to both magnetic field and velocity of charged particle and the charge moves in a curved path.

F = q v × B

The direction of the magnetic force is given by the right hand:

The fingers should be point in the direction of the magnetic field. Then the direction of palm gives the direction of magnetic force acting on positive charge and back of hand gives direction for magnetic force for negative charge.

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Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien
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The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
  • The height at which the block stops rising is approximately 1.1415 m
  • The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

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5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

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From the above solution for the height, the length of the incline is he

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  • Length of the incline, x_{max} = 1.536 m

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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
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Put the value into the formula

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\dfrac{1}{f}=\dfrac{9}{40}

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