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oksian1 [2.3K]
2 years ago
14

Hurryyyyyy When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand p

oints in the direction that the charge is moving? palm
fingers
back
thumb
Physics
2 answers:
masya89 [10]2 years ago
6 0
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.

</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
Ronch [10]2 years ago
3 0

Answer: The fingers.

Explanation: The magnetic force on a charge is given by the equation:

F = q(VxB)

where q is the charge of the particle, V is the velocity of the particle and B is the magnetic field.

If you use the right-hand method, you need to point with the tip of your fingers to the direction of the velocity, face the palm of your hand in the direction of the magnetic field, and now your thumb will point in the direction of the magnetic force.

So the part of the hand that points in the direction that the charge is moving is the second option, "fingers"

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The speed of sound in air changes with the temperature. When the temperature T is 32 degrees Fahrenheit, the speed S of sound is
dezoksy [38]

Answer and Explanation:

A. We have temperature t = 32

Speed of sound, s = 1087.5

As t increases by 1⁰f speed increases by 1.2

So that

S = 1088.6

T= 33⁰f

We have 2 equations

1087.5 = k(32) + c

1088.6 = k(33) + c

Subtracting both equations

(33-32)k = 1088.6-1087.5

K = 1.1

b.). S = kT + c

1087.5 = 32(1.1) + c

Such that

C = 1052.3

Therefore

S = 1.1(t) + 1052.3

C.). S = 1.1t + 1052.3

We make t subject of the formula

T = s/1.1 - 1052.3/1.1

T = 0.90(s) - 956.3

D. This means that We have temperature to rise by 0.90 whenever speed is increased

8 0
2 years ago
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a
bekas [8.4K]

Answer:

A) 12.08 m/s

B) 19.39 m/s

Explanation:

A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:

mg(sinθ) – F_k = ma

Where; F_k is frictional force due to kinetic friction given by the formula;

F_k = (μ_k) × F_n

F_n is normal force given by mgcosθ

Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)

a = -3.71 m/s²

We are told that distance d = 24.0 m and v_o = 18 m/s

Using newton's 3rd equation of motion, we have;

v = √(v_o² + 2ad)

v = √(18² + (2 × -3.71 × 24))

v = 12.08 m/s

B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

Using newton's 3rd equation of motion, we have;

v = √(v_o + 2ad)

v = √(18² + (2 × 1.08 × 24))

v = 19.39 m/s

6 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
lord [1]

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

5 0
2 years ago
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

7 0
2 years ago
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