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.
Hello there! Quantitive data has to do with measurements that can be shown with numbers. Examples of this are things like your height and the length of your arms. With that alone, A and B are eliminated, because those answer choices make no sense. They can't be expressed by numbers and you can't measure colors or odors mathematically. Volume is a way to measure something that CAN be written down by numbers. D is the only answer choice that fits the definition of quantitive data. The answer is D: volume.
Answer:
changing the polarity or direction of the battery changes the sign of the voltage and the current
Explanation:
The sign of current and voltage are due to established conventions.
The way that a DC circuit with negative current values is by changing the polarity of the power source or by inverting the battery, this creates that the electrons move in the opposite direction
Changing the battery also changes the direction of the power difference, since the potential from positive to negative, in most cases negative is assigned a potential of zero volts
In summary, changing the polarity or direction of the battery changes the sign of the voltage and the current
The answer is: "
44 
km " ;
or; write as: "
44.333 km " .
___________________________________________________________Explanation:___________________________________________________________(70 km + 63 km) ÷ (2 + 1 ) = 133 km ÷ 3 = "
44 km " ;
or; write as: "
44.333 km " .
___________________________________________________________
Answer:
V = 11.83 m/s
Explanation:
Given the following data;
Mass = 2000 kg
Force = 10000N
Distance = 14 m
To find the final velocity of the car;
First of all, we would determine the acceleration of the car;
Acceleration = force/mass
Acceleration = 10000/2000
Acceleration = 5 m/s²
Next, we would use the third equation of motion to find the final velocity;
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
V² = 0² + 2*5*14
V² = 0 + 140
V = √140
V = 11.83 m/s
