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bogdanovich [222]
3 years ago
5

If a positively charged particle moves into a magnetic field traveling in a straight line, how would you expect it's motion to c

hange
Physics
2 answers:
MrMuchimi3 years ago
6 0
It would begin to curve toward the magnetic pull
Aleksandr-060686 [28]3 years ago
4 0

Explanation:

If a positively charged particle moves into a magnetic field traveling in a straight line then there will be Lorentz force which acts on the positively charged particle.

The force exerts on a moving charge in a magnetic field. This force is Lorentz force.

The expression for Lorentz force is as follows;

F=q(v\times B)

F=qvBsin\theta

Here, q is the charge, v is the velocity, \theta is the angle between velocity and the magnetic field and B is the magnetic field.

There will be two conditions.

(1) If the positively charged particle moves parallel to the magnetic field then there will be no force on this particle. There is no deviation from the straight line. As the Lorentz force depends on the cross product of velocity and magnetic field. Sine 180 degree or sin 0 degree is zero in this case.

(2) If the positively charged particle moves perpendicular to the magnetic field then it follows the curved path until it completes the circle. The force is perpendicular to the direction of motion of the charged particle.

The charged particle will move in a circular motion.

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Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.
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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A
Snezhnost [94]

Answer:

9.96x10^-20 kg-m/s

Explanation:

Momentum p is the product of mass and velocity, i.e

P = mv

Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their production in standard alpha radioactive decay, alpha particles generally have a kinetic energy of about 5 MeV, and a velocity in the vicinity of 5% the speed of light.

From this we calculate the speed as

v = 5% 0f 3x10^8 m/s (speed of light)

v = 1.5x10^7 m/s

The mass of an alpha particle is approximately 6.64×10−27 kg

Therefore,

P = 1.5x10^7 x 6.64×10^−27

P = 9.96x10^-20 kg-m/s

8 0
2 years ago
A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist
galina1969 [7]

Answer:

0.04225 Nm

Explanation:

N = Force applied = 5 N

\mu = Coefficient of static friction = 0.65

d = Diameter of knob = 1.3 cm

r = Radius of knob = \frac{d}{2}=\frac{1.3}{2}=0.65\ cm

Force is given by

F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N

When we multiply force and radius we get torque

Torque on thumb

\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm

Torque on forefinger

\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm

The total torque is given by

\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm

The most torque that exerted on the knob is 0.04225 Nm

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2 years ago
When working with a powered instrument, the swirling effect produced by the water stream within the confined space of a periodon
Lina20 [59]

Answer:

Acoustic microstreaming

Explanation:

Acoustic microstreaming is the swirling effect produced by water stream confined in a spaced of a periodontal pocket.

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