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3 years ago

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A particle of charge q moving through a magnetic field B experiences the force where v is the velocity of the particle. Show tha

t: (i) The force does no work in moving the particle along its path from t = 0 to t=T. (i1) The speed v(t) Iv(O)ll of the particle is constant in time e particle 1s constant in time

1 answer:

MrRa [10]3 years ago

3 0

Answer:

dx/Dt x B . x =0

Explanation:

Let's calculate the work and the magnetic force, the expression for magnetic force is

F = qv x B

Bold indicate vector quantities, the expression for the job is

W = F. X

Let's replace in this equation

W = q v x B . X

The definition of speed is

v = dX / dt

With what work is left

W = q dX / dt x B . X

As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero

Second part

The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.

Let's calculate the magnitudes of speed (speed)

F = qv B sin θ

F = ma

q v B sin θ = ma

a = qvB / m senT

This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.

| v | = q vB/m sin θ

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Talia is on a trip with some friends. In the first 2 hours , they travel 100 miles. Then they hit traffic and go only 30 miles i

jeka94

Answer:

51 miles/hour

Explanation:

<em>Talia is on a road trip with some friends. in the first 2 hours, they travel 100 miles. then they hit traffic and go only 30 miles in the next hour. the last hour of their trip, they drive 75 miles. calculate the average speed of talia’s car during the trip. give your answer to the nearest whole number.</em>

<em />

<em>The average speed would be 51 miles/hour.</em>

<u>The average speed of a moving object is defined as the total distance traveled relative to the total time taken fro the journey. Mathematically, it is given as:</u>

total distance traveled/total time taken.

In this case;

Total distance traveled = 100 + 30 + 75 = 205 miles

Total time taken = 2 + 1 + 1 = 4 hours

Therefore;

Average speed = 205/4 = 51.25 miles/hour

<em>To the nearest whole number = 51 miles/hour</em>

5 0

3 years ago

If you want to study how energetic waves affect matter, you should study waves with a _______.

tankabanditka [31]

Higher Frequency and shorter wavelength. The energy of light waves increases when there is an increasing frequency and a higher frequency means there are shorter wavelengths. The equation lambda=c/f where lambda is wavelength and f is frequency and c is the speed of the wave.

4 0

3 years ago

Is shooting a a basketball force

lyudmila [28]

Answer:

yes

Explanation:

because when you shot the ball your applying force to it

3 0

3 years ago

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi

Kipish [7]

Answer:

A) pbin = 1.535 Kgm/s (+)

B) pbf = 1.696 Kgm/s (-)

C) Δp = 3.3925 Kgm/s

D) Δvr = 10.249 m/s

Explanation:

Given

Mass of the ball: m = 57.5 g = 0.0575 Kg

Initial speed of the ball: vbi = 26.7 m/s

Mass of the racket: M = 331 g = 0.331 Kg

Final speed of the ball: vbf = 29.5 m/s

A) We use the formula

pbin = m*vbi = 0.0575 Kg*26.7 m/s = 1.535 Kgm/s (+)

B) pbf = m*vbf = 0.0575 Kg*29.5 m/s = 1.696 Kgm/s (-)

C) We use the equation

Δp = pbf - pbin = 1.696 Kgm/s - (-1.535 Kgm/s) = 3.3925 Kgm/s

D) Knowing that

Δp = 3.3925 Kgm/s

we can say that

Δp = M*Δvr

⇒ Δvr = Δp / M

⇒ Δvr = 3.3925 Kgm/s / 0.331 Kg

⇒ Δvr = 10.249 m/s

8 0

3 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

nlexa [21]

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$ . The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$ . This is also the average velocity of the second leg.
The total time of this second leg is $t_2=1.60~min = 96~s$ . The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$ . Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$ . Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$

8 0

3 years ago