Borrowing from the tradition of the ancient Greeks, the first ones
who tried to describe the brightness of stars with numbers ...
6 magnitudes = a ratio of 100 .
In more modern notation . . .
1 magnitude = ratio of (100) ^ (1/6)
= about 2.154...
9.6 magnitudes = (100) ^ (9.6/6)
= 100¹·⁶
= 1,584.89 (rounded)
P= 10m = 10 x 9.10 = 91 N
Work is done lifting a 9.10-kg box straight up onto a shelf that is 1.80 m high :
A= Ph = 91 x 1.80 = 163.8 J
ok done. Thank to me :>
Answer:
11,100 N
Explanation:
Draw the car on the incline. Label the weight and normal forces. Call the direction normal to the incline y, and the direction parallel to the incline x.
If it helps, rotate the incline so that the surface of the incline is flat.
Use trigonometry to determine the components of the weight force. The y component of the weight is:
Wᵧ = mg cos θ
Wᵧ = (1150 kg) (9.8 m/s²) (cos 8.70°)
Wᵧ = 11,100 N
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
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.