Current doesn't follow parallelogram rule, but it has direction. Is it neither vector nor scalar? Doesn't scalar have no direction?

Current doesn't follow parallelogram rule, but it has direction. Is it neither vector nor scalar? Doesn't scalar have no direction?

Electric current is really special. In high school, we described it like this: "it is a scalar with direction, and its direction changes with the direction and shape of the conductor."
Be sure to remember that it is scalar when taking an exam. This is a small test point
Hope to help you ~! Do not understand Baidu Hi I ~!

In the experiment of "parallelogram rule of verification force", one end of the rubber band is fixed at point a, and the other end is pulled to point O by two spring scales. The readings of the two spring scales are F1 and F2 respectively. The angle between the direction of the two strings and the extension line of the rubber band is α 1 and α 2 respectively, as shown in the figure
A. As long as the position of o point remains unchanged, then the resultant force remains unchanged. B. in the experiment, the resultant force F obtained by using the parallelogram rule must be along the direction of OA straight line. C. if the position of 0 point and the angle α 1 remain unchanged, F1 increases and F2 may decrease. D. the resultant force F must be greater than F1 or F2

A. If the position of o point remains unchanged, the shape of rubber band is the same, so the resultant force remains unchanged, so a is correct; B. because there are errors in the experiment, it is impossible to make f and f 'coincide strictly, so the resultant force F obtained by using the parallelogram rule in the experiment is not along the direction of OA straight line, so B is wrong; C. according to the fixed value of parallelogram, if the position and angle α 1 of 0 point remain unchanged, F1 increases If α 2 also increases, F2 may decrease, so C is correct; D. because the resultant force and component force satisfy the parallelogram rule, the geometric relationship shows that the resultant force may be less than two components, so D is wrong; so AC

Experimental procedure of parallelogram rule for verifying force

The one above is too professional. I'm afraid it's not what you want. 1. Fix one end of the rubber band at point a on the board. 2. Knot two strings at the other end of the rubber band. Pull the rubber band at an angle with two spring scales through the string. The rubber band extends to make the node extend to point O (as shown in the figure)

How to find the indefinite integral of the - x power of E?
And how to find the indefinite integral of the absolute value of SiNx?
I only know how to find SiNx and - SiNx by classification
We get - cosx + C and cosx + C, but the answer is - cosx + 4K and cosx + 4K + 2

For indefinite integral:
(1).∫e^(-x)dx
The original formula = - ∫ d [e ^ (- x)] = - e ^ (- x) + C
(2).∫∣sinx∣dx
When 2K π≤ x ≤ (2k + 1) π, SiNx ≥ 0, then ∫ ∣ SiNx ∣ DX = ∫ sinxdx = - cosx + C;
When (2k + 1) π ≤ x ≤ 2 (K + 1) π, SiNx ≤ 0, then ∫ ∣ SiNx ∣ DX = - ∫ sinxdx = cosx + C;
The following integral constant C, written as 4K or 4K + 2 in your answer, may have its special need to solve the problem, it doesn't matter

What is the indefinite integral of Sint / T?
Can you give the specific operation process?

This function is not integrable, but its original function exists, but it can't be expressed by elementary function
Traditionally, if the original function of a given continuous function can be expressed by an elementary function, it is said that the function is a "product function", otherwise it is said that it is a "product function". For example, several integrals listed below are "product function", but these integrals play an important role in probability theory, number theory, optics, Fourier analysis and other fields
∫ e ^ (- x * x) DX, ∫ (SiNx) / xdx (integral in the title), ∫ 1 / (LNX) DX, ∫ sin (x * x) DX
∫ (a * a * SiNx * SiNx + b * b * cosx * cosx) ^ (1 / 2) DX (a * a is not equal to b * b)
So, don't worry about integrating the objective function

Solving indefinite integral: ∫ (3sint + (1 / Sint ^ 2t)) DT

｜ab-2｜+｜b-1｜=0,
So ab-2 = B-1 = 0
So B = 1
ab=2
a=2/b=2
So the original formula = 1 / 1 * 2 + 1 / 2 * 3 + +1/2013*2014
=1-1/2+1/2-1/3+…… +1/2013-1/2014
=1-1/2014
=2013/2014

What is the indefinite integral of cos θ 4 power

Cos ^ 4 θ = (COS ^ 2 θ) ^ 2 = (1 / 4) (1 + cos 2 θ) ^ 2 = (3 / 8) + (1 / 2) cos 2 θ + (1 / 8) cos 4 θ. Then the integral is: ∫ cos ^ 4 θ D θ = (3 θ / 8) + (1 / 4) ∫ cos 2 θ D (2 θ) + (1 / 32) ∫ cos 4 θ D (4 θ) = (3 θ / 8) + (1 / 4) sin 2 θ + (1 / 32) sin 4 θ + C

How to solve the indefinite integral of the sixth power of COS!
What's its definite integral from 0 to 2 / π?

so easy let me teach you.
cos⁶x
= (cos²x)³
= [(1 + cos2x)/2]³
= (1/8)(1 + cos2x)³
= (1/8)(1 + 3cos2x + 3cos²2x + cos³2x)
= 1/8 + (3/8)cos2x + (3/8)cos²2x + (1/8)cos²2xcos2x
= 1/8 + (3/8)cos2x + (3/8)(1 + cos4x)/2 + (1/8)(1 + cos4x)/2 · cos2x
= 1/8 + (3/8)cos2x + 3/16 + (3/16)cos4x + (1/16)cos2x + (1/16)cos4xcos2x
= 5/16 + (7/16)cos2x + (3/16)cos4x + (1/16)(1/2)(cos6x + cos2x)
= 5/16 + (15/32)cos2x + (3/16)cos4x + (1/32)cos6x
∴∫ cos⁶x dx
= 5x/16 + (15/32)(1/2)sin2x + (3/16)(1/4)sin4x + (1/32)(1/6)sin6x + C
= 5x/16 + (15/64)sin2x + (3/64)sin4x + (1/192)sin6x + C
The method used upstairs is right, but the calculation is wrong. It should be as follows
∫ cos⁶x dx
= (1/8)∫ (1 + 3cos2x + 3cos²2x + cos³2x) dx
= (1/8)∫ dx + (3/8)∫ cos2x dx + (3/8)∫ cos²2x dx + (1/8)∫ cos²2x cos2x dx
= x/8 + (3/8)(1/2)sin2x + (3/8)(1/2)∫ (1 + cos4x) dx + (1/8)(1/2)∫ cos²2x dsin2x
= x/8 + (3/16)sin2x + (3/16)(x + 1/4 · sin4x) + (1/16)∫ (1 - sin²2x) dsin2x
= x/8 + (3/16)sin2x + 3x/16 + (3/64)sin4x + (1/16)[sin2x - (sin³2x)/3] + C
= 5x/16 + (1/4)sin2x + (3/64)sin4x - (1/48)sin³2x + C
The mistake is step 4 (1 / 16) ∫ (1 - Sin & # 178; 2x) dsin2x = (1 / 16) (sin2x - (Sin & # 179; 2x) / 3) ≠ (1 / 16) (x - (Sin & # 179; 2x) / 3)
This integral can be expressed by a special formula from 0 to π / 2
∫(0→π/2) cos⁶x dx
= (6 - 1)!/6! · π/2
= 5/6 · 3/4 · 1/2 · π/2
= 5π/32
For the formula such as ∫ (0 → π / 2) Sin & # 8319; DX = ∫ (0 → π / 2) cos & # 8319; X DX, n > 1
When n is odd
= (n - 1)!/n! = (n - 1)/n · (n - 3)/(n - 2) · (n - 5)/(n - 4) · ... · 3/4 · 1/2
When n is even
=(n - 1)! / N! · π / 2 = (n - 1) / N · (n - 3) / (n - 2) · (n - 5) / (n - 4) ·... · 3 / 4 · 1 / 2 · π / 2, more π / 2

How to calculate (Sint + cost) / (cost Sint), equal to tan (T + π / 4)

(sint + cost) / (cost-sint)
= [(sint + cost)/√2] / [(cost - sint)/√2]
= (sint cosπ/4 + cost sinπ/4) / (cost cosπ/4 - sint sinπ/4)
= sin(t+π/4) / cos(t+π/4)
= tan(t+π/4)

Find the maximum square of y = cosx times SiNx + SiNx

y=cosx*sinx+sinx*sinx?
y=1/2*sin(2x)-1/2*cos(2x)+1/2
y'=cos(2x)+sin(2x)
Let y '= 0
cos(2x)=-sin(2x)=±√2/2
Get the maximum value of 1 / 2 + √ 2 / 2
Minimum 1 / 2 - √ 2 / 2