Ask question

Ask question

All categories

2 years ago

3

Slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.0400

kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest.

1 answer:

abruzzese [7]2 years ago

7 0

Answer:

The linear speed is 1.33 m/s.

Explanation:

Given that,

Length of rod = 90.0 cm

Mass of slender rod = 0.120 kg

Mass of small sphere = 0.0200 kg

Mass of another small sphere = 0.0400 kg

Suppose, we need to find the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

We need to calculate the the change in potential of the complete system

Using formula of change in potential

m₂ and m₃ are the masses at the rod ends.

The rod center of mass neither gains nor loses potential

$\Delta U=m_{2}gy_{2}+m_{3}gy_{3}$

Put the value into the formula

$\Delta U=0.0400\times9.8\times(-45\times10^{-2})+0.0200\times9.8\times45\times10^{-2}$

$\Delta U=-0.0882\ N-m$

We need to calculate the moment of inertia of the rod

Using formula of moment of inertia of the rod

$I_{1}=2\rho \int_{0}^{r}(r^2 dr)$

Put the value into the formula

$I_{1}=2\times\dfrac{0.120}{90\times10^{-2}} \int_{0}^{0.45}(r^2 dr)$

$I_{1}=2\times\dfrac{0.120}{90\times10^{-2}}\times(\dfrac{(0.45)^3}{3}-0)$

$I_{1}=0.0081\ kg-m^2$

We need to calculate the moment of inertia of the end masses

Using formula of moment of inertia

$I_{2+3}=\sum mr^2$

Put the value into the formula

$I_{2+3}=(0.0400+0.0200)\times0.45^2$

$I_{2+3}=0.01215\ kg-m^2$

We need to calculate the change in potential energy to the system kinetic energy

Using formula of kinetic energy

$\Delta U=K.E$

$\Delta U=\dfrac{1}{2}(I_{1}+I_{2+3})\omega^2$

Put the value into the formula

$0.0882=\dfrac{1}{2}(0.0081+0.01215)\omega^2$

$\omega^2=\dfrac{2\times0.0882}{0.02025}$

$\omega=\sqrt{\dfrac{2\times0.0882}{0.02025}}$

$\omega=2.95\ rad/s$

We need to calculate the linear speed

Using formula of linear speed

$v = r\omega$

Put the value into the formula

$v=0.45\times2.95$

$v=1.33\ m/s$

Hence, The linear speed is 1.33 m/s.

You might be interested in

A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro

LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock = 0.22kg

Initial velocity = 20m/s

Distance moved = 10m

Unknown:

Initial kinetic energy of the rock = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2}$ x 0.22 x 20² = 44J

6 0

1 year ago

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit

melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0

2 years ago

A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca

xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance = 4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm$

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2$

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

8 0

2 years ago

This is really urgent

hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0

2 years ago

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

tekilochka [14]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

4 0

2 years ago