1. How do you describe the path that the light passes through when it travels?
2. What happens during dispersion of light?
3. What are the different colors of light?
4. Similarly, how are rainbows formed up in the sky?


pls help me if u can

For the electric field is measured for points at distances r,Electric field  is mathematically given as

E=19.9*10^{-5}c/m^3

Generally, the equation for the  electric field uniform charged is mathematically given as

[tex]E=\frac{lr}{3e0}[/tex]

Therefore

E=lr/3E0=3Ee0/r

Therefore

E=3*6*10^4*8.85*10^{-12}

E=19.9*10^{-5}c/m^3

In conclusion, E=19.9*10^{-5}c/m^3

E=19.9*10^{-5}c/m^3

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CQ

Complete question attached below

What's the relationship

[tex]\\ \rm\Rrightarrow T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

[tex]\\ \rm\Rrightarrow T^2g=4\pi ^2\ell[/tex]

[tex]\\ \rm\Rrightarrow\ell=\dfrac{T^2g}{4\pi^2}[/tex]

4A can be correct

The correct option is: 2A cm

In a simple harmonic motion, the particle moves back and forth around a mean position. The distance covered by the particle in one time period depends on the amplitude of the oscillation.

The time period (T) of a simple harmonic motion can be expressed in terms of the frequency (f) as:

T = 1/f

The frequency (f) of a simple harmonic motion can be related to the angular frequency (ω) as:

ω = 2πf

Therefore, the time period (T) can also be expressed in terms of the angular frequency (ω) as:

T = 2π/ω

The displacement of the particle at any time (t) during the simple harmonic motion can be expressed as:

x = A cos(ωt)

where x is the displacement, A is the amplitude, and cos(ωt) is the cosine function.

The particle moves from its initial position to its maximum displacement (+A), then back to its initial position, and finally to its maximum displacement in the opposite direction (-A), before returning to its initial position again. Therefore, the distance covered by the particle in one time period is twice the amplitude (2A).

Therefore, the correct option is:

2A cm

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The half life is the time taken for half the number of radioactive atoms originally present to remain. This is  the time that corresponds to exactly half of the original number of radioactive isotopes remaining in the system.

The half life is the time taken for half the number of radioactive atoms originally present to remain. We can plot the number of atoms against the time taken in a graph.

We can obtain the half life of the radioactive isotope from the graph by noting the time that corresponds to exactly half of the original number of radioactive isotopes remaining in the system.

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Answer:

Explanation:

Formula

MA = length of ramp / height

Givens

ramp length = 9 feet

height = 3 feet

Solution

MA = 9 / 3

MA = 3

Answer Mechanical advantage = 3

Answer:

Rolling friction

Explanation:

Option B is correct

Answer:

A,B,C

Explanation:

because that is true

The maximum value of the current depend on the frequency of the ac voltage source because it increases as the frequency increases.

The peak current of an ac source is the amplitude current flowing in the ac circuit.

The magnitude of the peak  current is calculated as follows;

I₀ = i/(sin ωt)

I₀ = i/(sin 2πft)

where;

i = I₀ x sin 2πft

From the equation above, as the frequency of the of the ac voltage source increases, the value of the ac current (i) increases.

Thus, the maximum value of the current depend on the frequency of the ac voltage source because it increases as the frequency increases.

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For E = 200 gpa and i = 65. 0(106) mm4,  the slope of end a of the cantilevered beam  is mathematically given as

A=0.0048rads

Generally, the equation for the   is mathematically given as

[tex]A=\frac{PL^2}{2EI}+\frac{ML}{EI}[/tex]

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion,  the slope is

A=0.0048rads

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The value of slope of end a of the cantilevered beam when E = 200 gpa and i = 65. 0(106) mm4 is 0.0048 rads.

The formula used to find the slope at the end of the cantilevered beam is,

[tex]A=\dfrac{PL^2}{2EI}+\dfrac{ML}{EI}[/tex]

Here, (M) is the momentum, (EI) is flexural rigidity, (P) is concentrated load and (L) is the length of the beam.

Concentrated load is 10 KN and uniform load is 3 kN/m. E = 200 gpa and i = 65. 0(106) mm4.

[tex]A=\dfrac{10+10^2+3^2}{2\times240\times10^9\times65\times10^6}+\dfrac{10+10^3\times3}{240\times10^9\times65\times10^{-6}}\\A=0.00288+0.00192A=0.0048 \rm\; rads[/tex]

Thus, the value of slope of end a of the cantilevered beam when E = 200 gpa and i = 65. 0(106) mm4 is 0.0048 rads.

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[tex]\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

[tex]\\ \rm\rightarrowtail T^2=4\pi^2\dfrac{l}{g}[/tex]

[tex]\\ \rm\rightarrowtail g=\dfrac{4\pi ^2\ell}{T^2}[/tex]

[tex]\\ \rm\rightarrowtail g=\left(\dfrac{2\pi\sqrt{l}}{T}\right)^2[/tex]

Answer:

30 cm to 1 mm

Explanation:

30 cm to 1 mm

The acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).

The acceleration of the mass down the plane is determined by applying the principle of conservation of angular momentum.

Fr = Iα

where;

(mg sinθ)r = Iα

(mg sinθ)r = I(ar)

(mg sinθ) = I(a)

[tex]a = \frac{mg \times sin(\theta)}{I} \\\\a = \frac{mg \times sin(\theta)}{3mr^2/4} \\\\a = \frac{4mg \times sin(\theta)}{3mr^2}[/tex]

Thus, the acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).

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Answer:

the correct one is A hope it helps

[tex]\text{Density,}\\\\~~~~~~~~\rho = \dfrac mV\\\\\\\implies V = \dfrac m{\rho}\\\\\\\implies V = \dfrac{7 \times 10^{-3}~ \text{kg}}{\tfrac{5.6 \times 10^{-3}~ \text{kg}}{ 10^{-6} ~\text{m}^3} }\\\\ \\\implies V = \dfrac{7 \times 10^{-3}~ \text{kg}}{5.6 \times 10^{-3} \times 10^6~~ \text{kg}~ \text{m}^{-3}}\\\\\\ \implies V = 1.25 \times 10^{-6}~ \text m^3\\\\\\\implies V =0.00000125~ \text m^3\\\\\text{Hence the volume of the coin is 0.00000125}~ \text m^3[/tex]

Explanation:

Nope, displacement is a vector quantity, whereas distance is the scalar (magnitude) of the total displacement.

Hope this clears things up :))

Answer:

Current for 12 ohms = 2.5

Explanation:

Ohm's Law:

Voltage = IR

Current = V/R

Resistance = V/I

Therefore, current = 30V/12 Ohms

30/12 = 2.5

So current for 12 ohms = 2.5

I hope this helps :>

Hi there!

We can use Ohm's law to solve:
[tex]V = IR[/tex]

V = Potential Difference (V)
I = Current (A)
R = Resistance (Ω)

We can rearrange the equation to solve for resistance:

[tex]R = \frac{V}{I}[/tex]

Plug in the given values to solve.  

[tex]R = \frac{10}{2} = \boxed{5 \Omega}[/tex]

Answer:

Explanation:

The picture is cut off so the wrecking ball and its path are not shown. It would be better if the whole picture is available.

Based on the written info, at the bottom of the path, the wrecking ball is moving at 10m/s.

The centripetal acceleration = velocity^2 / radius

= 10^2 / 25

= 100 / 25

= 4m/s^2

The centripetal force = mass * centripetal acceleration

= 600*4

= 2400N

As the ball is moving horizontally at the bottom, net vertical force is zero.

So Tension force in cable = Wrecking ball weight + Centripetal force

= 600*g + 2400

= 6000 + 2400; assuming g is 10m/s^2

= 8400N

Answer:

It becomes less heavy

Explanation:

Hopefully it helps!

Answer:

-18,750N

Explanation:

90x1000

-------------

3600

u=25m/s

v=0m/s

v^2=u^2+2as

v=0

=625+(2×a×20)

625+40a

-625=40a

a= -15.625 m/s^2

F=ma

=(1200)(-15.625)

= -18,750N

Answer:

Temporary? You have two options

If you have a positively charged object, bring it close to the LEFT side of the ball (not touching it!).

If you have a negatively charged object, bring it close to the RIGHT side of the ball (again, not in contact).

Either way, the electron cloud will migrate towards the right side of the of your sphere.

Another option is to have a capacitor large enough to house the sphere inside it, and polarize it by attaching it to a DC source making sure that the negative plate of it is to the right of the sphere.

Answer:

The inverse square law proposed by Newton suggests that the force of gravity acting between any two objects is inversely proportional to the square of the separation distance between the object's centers.

Explanation:

The electric potential energy of the charge will be =3.5*10^8 J  

Electric potential energy is the energy that is needed to move a charge against an electric field.

F=Force experienced by the charge = 3.6*10^-4 N

q1 = magnitude of charge producing the electric field

q2 = magnitude of charge experiencing the electric force

r1 = distance between the two charges

Electric force experienced by the charge is given using coulomb's law as

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

[tex]3.6\times 10^{-4}=\dfrac{9\times 10^9 q_1q_2}{(9.8\times 10^{-5})^2}[/tex]

[tex]q_1q_2=3.84\times 10^{-22}[/tex]

Electric potential energy of the charge can be given as

[tex]U=\dfrac{kq_1q_2}{r}[/tex]

[tex]U=\dfrac{(9\times 10^9)q_1q_2}{(9.8\times 10^{-5})^2}[/tex]

[tex]U=3.5\times 10^{-8}\ J[/tex]

Thus Electric potential energy of the charge can be given as [tex]U=3.5\times 10^{-8}\ J[/tex]

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Answer:

An electron is a component of the broader term, electricity. Generally electricity refers to the control, manipulation, and use of the movement of electrons to perform some kind of work (motion/rotation, light, heat, etc…).

Explanation:

Or

put this

The study of electrons as it relates to their use in electric devices is electricity.

But I will go with the first one

Answer:

What is your name?

Explanation:

no bro I don't know I am New of this app so I don't know

Answer:

A

Explanation:

i think its A

Answer:

An isotope is one of two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behavior but with different atomic masses and physical properties. In medicine, for example, cobalt-60 is extensively employed as a radiation source to arrest the development of cancer. Other radioactive isotopes are used as tracers for diagnostic purposes as well as in research on metabolic processes.

Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day!

Those isotopes which are unstable due to the presence of extra neutrons in their nuclei and emit various types of radiations are called radioactive isotopes. The radiations emitted are in the form of alpha particles, beta particles and gamma rays.

For eg :- carbon,hydrogen and chlorine.

Volume:-

[tex]\\ \sf\rightarrowtail \dfrac{4}{3}\pi r^3[/tex]

[tex]\\ \sf\rightarrowtail \dfrac{4}{3}\pi (0.18)^3[/tex]

[tex]\\ \sf\rightarrowtail \dfrac{4}{3}\pi (0.005832)[/tex]

[tex]\\ \sf\rightarrowtail 0.024m^3[/tex]

Density:-

[tex]\\ \sf\rightarrowtail \dfrac{Mass}{Volume}[/tex]

[tex]\\ \sf\rightarrowtail \dfrac{190}{0.024}[/tex]

[tex]\\ \sf\rightarrowtail 7916.67kg/m^3[/tex]

The total moment of inertia about an axis is : [tex]L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}[/tex] for a ring of mass m and radius straight r attached to a thin rod.

Given data:

mass of ring --> m

radius of ring --> r

mass of rod --> M

Length of rod ---> L ( 2 * radius )

Total Moment of Inertia about an axis = Irod  +  Iring

where : Irod = moment of inertia of rod,  Iring = moment of inertia of ring

Irod = ML² / 3

Iring = 2mr² / 5

moment of inertia around an axis by Iring = I

where ;  I = 2mr² / 5  + ML²   according to parallel axis theorem

Hence the Total moment of Inertia about an axis is :

Itotal =  2mr²/5  +  ML²  +  ML² / 3

        = [tex]L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}[/tex]

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For a ring of mass m and radius straight r connected to a thin rod, the total moment of inertia about an axis is [tex]\rm I= L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}[/tex].

The sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation expresses a body's tendency to resist angular acceleration.

The given data in the problem is;

m is the ring's mass.

r is the radius of a ring.

L is the rod length = 2r

The total Moment of Inertia about an axis;

[tex]\rm I = I_{rod}+I_{ring} \\\\ \rm I =\frac{ML^2}{3} +\frac{2mr^2}{5}[/tex]

According to the parallel axis theorem,

[tex]\rm I = \frac{2mr^2}{5} + ML^2[/tex]

The total Moment of Inertia about an axis will be;

[tex]\rm I_{total} = \frac{ML^2}{3} +\frac{2mr^2}{5}+ML^2 \\\\ \rm I_{total} =L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}[/tex]

Hence for a ring of mass m and radius straight r connected to a thin rod, the total moment of inertia about an axis is [tex]\rm I= L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}[/tex].

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Answer: KE = 172.872 J

PE = 307.328 J

Explanation:

For KE: acceleration = change in velocity / time → 9.8 = final velocity - 0 (because initial velocity is zero) / 6s → final velocity after 6s = 58.8 m/s.

Now, plug this into: KE = 1/2(m)(v^2) → KE = 1/2(.1)(58.8^2) = 172.872 J.

For PE: Find the velocity after 10 s: acceleration = change in velocity / time → 9.8 = final velocity - 0 (because initial velocity is zero) / 10s → final velocity after 10s = 98 m/s.

Now, find the total height: (v final)^2 = (v initial)^2 + 2*a*d → 98^2 = 0^2 + 2*9.8*d → d = total height = 490 m.

Now, find the change in height after 6 s: (v final)^2 = (v initial)^2 + 2*a*d → 58.8^2 = 0^2 + 2*9.8*d → d = change in height after 6 s → 176.4 m

Now, subtract the change in height from the total height to get its height after 6s: 490-176.4=313.6m

PE = m*g*h → Pe=.1*9.8*313.6 = 307.328 J :)