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olga2289 [7]
3 years ago
9

Two spheres A and B of negligible dimensions and masses 1 kg and √3 kg respectively, are supported on the smooth circular surfac

e, fixed to the ground with a centre O and radius of 0.1m. The spheres are joined by the cord shown in length π/20 m; determine the angles α and β corresponding to the position of equilibrium of the spheres with respect to the vertical passing through O.

Physics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

α = π/3

β = π/6

Explanation:

Use arc length equation to find the sum of the angles.

s = rθ

π/20 m = (0.1 m) (α + β)

π/2 = α + β

Draw a free body diagram for each sphere.  Both spheres have three forces acting on them:

Weight force mg pulling down,

Normal force N pushing perpendicular to the surface,

and tension force T pulling tangential to the surface.

Sum of forces on A in the tangential direction:

∑F = ma

T − m₁g sin α = 0

T = m₁g sin α

Sum of forces on B in the tangential direction:

∑F = ma

T − m₂g sin β = 0

T = m₂g sin β

Substituting:

m₁g sin α = m₂g sin β

m₁ sin α = m₂ sin β

(1 kg) sin α = (√3 kg) sin (π/2 − α)

1 sin α = √3 cos α

tan α = √3

α = π/3

β = π/6

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Explanation:

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As the frequency of a wave increases the wavelength.
zaharov [31]

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decreases

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Read 2 more answers
An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached
Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

6 0
3 years ago
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and
Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

         W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

         Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         mg h = K

to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

            h = L (1-cos 35)

we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

7 0
2 years ago
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