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Oksi-84 [34.3K]

2 years ago

6

A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizonta

l circle so that the thread describes a cone. Calculate the magnitude of the angular momentum of the bob about a vertical axis through the supporting point. The acceleration of gravity is 9.8 m/s2. Answer in units of kg · m2/s

1 answer:

shusha [124]2 years ago

8 0

Answer:

19.99 kg m²/s

Explanation:

Angular Momentum (L) is defined as the product of the moment of Inertia (I) and angular velocity (w)

L = m r × v.

r and v are perpendicular to each other,

where r = lsinθ.

l = 2.4 m

θ= 34°

g = 9.8 m/s² and m = 5 kg

resolving using newtons second law in the vertical and horizontal components.

T cos θ − m g = 0

T sin θ − mw² lsin θ = 0

where T is the force with which the wire acts on the bob

w = √g / lcosθ

= √ 9.8 / 2.4 ×cos 34

= 2.2193 rad/s

the angular momentum L = mr× v

= mw (lsin θ)²

= 5 × 2.2193 (2.4 ×sin 34°)²

=19.99 kg m²/s

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Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k

Explanation:

Thermal comductivity K = QL/A∆T

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L= Distance between two isothermal planes = 0.740mm

A= Area of the surface in square metres = 2m^2

∆T= Temperature change = (37-30) °C.

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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

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2 years ago

There is an electromagnetic wave traveling in the -z direction in a standard right-handed coordinate system. What is the directi

wlad13 [49]

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (<em>in this case the -z direction</em>) and the curling of the fingers point in the direction of the magnetic field B→ (<em>in this case the +x direction</em>), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

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2 years ago

The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this

konstantin123 [22]

Answer:

$3.81972\times 10^{-7}\ A$

Explanation:

B = Magnetic field = $3\times 10^{-8}\ G$

d = Diameter of loop = 16 cm

r = Radius = $\frac{d}{2}=\frac{16}{2}=8\ cm$

i = Current

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

The magnetic field of a loop is given by

$B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A$

The current needed to produce such a field at the center of the loop is $3.81972\times 10^{-7}\ A$

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2 years ago

A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses

gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

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