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Roman55 [17]
3 years ago
13

What is the difference between the equilibrium vector and the resultant vector

Physics
1 answer:
Tresset [83]3 years ago
8 0

the difference between a resultant and equilibrant vector is that resultant vector is a direct quantity, one with both magnitude and direction, while the equilibrant vector is a force equal to, but opposite of, the resultant sum of vector forces, that force which balances other forces.

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Light traveling in water (n = 1.33) into an unknown medium.If rhe angles if incidence and refraction are 40 degrees and 25 degre
Bess [88]
40 degreased and 25 dregrees=65 degrees‍♀️
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3 years ago
How do you know that potassium an alkali metal is highly reactive
Pavel [41]

Potassium is in the most reactive group of elements, the alkali metals, but it's not the most reactive metal within the group. The alkali metals, Group 1A, are the most reactive metals because they have one valence or outer electron. They lose this electron very easily, forming ions with a charge of +1.

7 0
3 years ago
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
3 years ago
How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
3 years ago
FILL IN THE BLANKS!
Aleksandr [31]
1. First blank is A. Conductors
Second blank is D. Insulators

2. C. Heat
8 0
3 years ago
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