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nadezda [96]
3 years ago
9

You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths

. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

The free end must be attached at a distance of 27.5 cm

Solution:

Mass, m = 141 kg

Length of the rods, L= 55 cm

Now,

As clear from fig. 1:

The free end of the rod 2 must be attached at:

F = 2 W

WL = W(55 - L)

2L = 55

L = 27.5 cm

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Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths f
Bogdan [553]

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t_{min =  ( m + 1/2) × λ/2n

we substitute

t_{min =  ( 0 + 1/2) × 711 /2(1.21)

t_{min = 0.5 × 711/2.42

t_{min = 0.5 × 293.80165

t_{min = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

8 0
3 years ago
A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

  • Disk radius R = 15 cm = 0.15 m
  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0
2 years ago
(a) Find the acceleration of B.<br>(b) Find the tensions, T1 and T2, in the strings.
Ann [662]

i believ that the answer would be

the acceleration of B is 0.2

6 0
3 years ago
What were the patterns that people saw in the sky long ago?
Lena [83]
Long time ago, people saw the constellations as patterns in the sky.  They names these patterns and tell stories about them. What people saw laong time ago are just mere patterns which forms animals and shapes.  We got the names of our constellations from the Greeks who named the constellations after the mythological heroes and mythological legends.
7 0
3 years ago
Read 2 more answers
A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the
UkoKoshka [18]

Answer:

 Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, KE = \frac{1}{2} mv^2, where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

  1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg

So, mass of bike = 38 kg.

3 0
3 years ago
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