Answer:
(a) 7.315 x 10^(-14) N
(b) - 7.315 x 10^(-14) N
Explanation:
As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,
(a) The electron has a velocity defined as: ![\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%282.4x10%5E%7B6%7D%20i%20%2B%203.6x10%5E%7B6%7D%20j%29%20%5Cfrac%7B%5Bm%5D%7D%7B%5Bs%5D%7D%5C%5C%5C%5C)
In respect to the magnetic field; ![\overrightarrow{B}=(0.027 i - 0.15 j) [T]](https://tex.z-dn.net/?f=%5Coverrightarrow%7BB%7D%3D%280.027%20i%20-%200.15%20j%29%20%5BT%5D)
The magnetic force can be written as;
![\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%20%3D%20q%28%5Coverrightarrow%7Bv%7D%20x%20%5Coverrightarrow%7BB%7D%29%5C%5C%20%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%20q%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2.4x10%5E%7B6%7D%263.6x10%5E%7B6%7D%260%5C%5C0.027%26-0.15%260%5Cend%7Barray%7D%5Cright%5D)
Bear in mind ![q =-1.6021x10^{-19} [C]](https://tex.z-dn.net/?f=q%20%3D-1.6021x10%5E%7B-19%7D%20%5BC%5D%20)
thus,
![\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%3D%20q%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2.4x10%5E%7B6%7D%263.6x10%5E%7B6%7D%260%5C%5C0.027%26-0.15%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%20q%282.4x10%5E%7B6%7D%2A%20%28-0.15%29-%20%280.027%2A3.6x10%5E%7B6%7D%29%29%5C%5C%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%20-1.6021x10%5E%7B-19%7D%20%5BC%5D%28-457200%29%20%5BT%5D%5Cfrac%7Bm%7D%7Bs%7D%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%287.3152x10%5E%7B-14%7D%29%20k%20%5B%5Cfrac%7BN%2Am%2Fs%7D%7BC%2Am%2Fs%7D%5D%5C%5C%5C%5C%7CF%7C%3D%20%5Csqrt%7B%20%287.3152x10%5E%7B-14%7D%29%5E%7B2%7D%5BN%5D%5E2%20%2Ak%5E%7B2%7D%7D%5C%5C%5C%5CF%3D7.3152x10%5E%7B-14%7D%20%5BN%5D)
Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, 
(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus
![\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%3D%20q%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2.4x10%5E%7B6%7D%263.6x10%5E%7B6%7D%260%5C%5C0.027%26-0.15%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%20q%282.4x10%5E%7B6%7D%2A%20%28-0.15%29-%20%280.027%2A3.6x10%5E%7B6%7D%29%29%5C%5C%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%201.6021x10%5E%7B-19%7D%20%5BC%5D%28-457200%29%20%5BT%5D%5Cfrac%7Bm%7D%7Bs%7D%5C%5C%5C%5C%5Coverrightarrow%7BF%7D%3D%28-7.3152x10%5E%7B-14%7D%29%20k%20%5B%5Cfrac%7BN%2Am%2Fs%7D%7BC%2Am%2Fs%7D%5D%5C%5C%5C%5C%7CF%7C%3D%20%5Csqrt%7B%20%28-7.3152x10%5E%7B-14%7D%29%5E%7B2%7D%5BN%5D%5E2%20%2Ak%5E%7B2%7D%7D%5C%5C%5C%5CF%3D-7.3152x10%5E%7B-14%7D%20%5BN%5D)
Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,
Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is 
Visible light waves are the type of electromagnetic waves that make up the colors of the rainbow, because the rainbow is visible to us.
-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is
Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>
<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.
After the jump:
-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.
His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.
But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.
Without Isaac, the boat's mass is 300 kg, so
(300 x speed) = 36 kg-m/s .
Divide each side by 300: speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================
Another way to do it . . . maybe easier . . . in the frame of the boat.
In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.
Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.
Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.
In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186
Divide each side by 300: speed = 186/300 = <em>0.62 m/s</em> <u>away</u> from the jump.
Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.
The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.
yay !
By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.
