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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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3 years ago
What is the speed of a car that traveled 500 meter in 30 seconds?
GenaCL600 [577]
<h2>Greetings!</h2>

To find speed, you need to remember the formula:

Speed = distance ÷ time

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3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
Read 2 more answers
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
3 years ago
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Svetach [21]

Answer:

F= 0.009 N

Explanation:

Given that

Charge ,q= 5.13 μC

Velocity ,V= 8.64 x 10⁶ m/s

Magnetic field , B = 1.99 x 10⁻⁴ T

The force on a charge q moving with velocity v is given as follows

F= q V B

Now by putting the values in the above equation we get

[tex]F= 5.13\times 10^{-6}\times 8.64\times 10^{6}\times 1.99\times 10^{-4}\ N [\tex]

F=0.00882 N

F= 0.009 N

Therefore the force on the particle will be 0.009 N.

5 0
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