Answer:
T = 1.12 10⁻⁷ s
Explanation:
This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.
All the charge dq is at a distance r
dE = k dq / r²
Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry
cos θ =[tex]\frac{dE_x}{dE}[/tex]
dEₓ = dE cos θ
cos θ = x / r
substituting
dEₓ = [tex]k \frac{dq}{r^2 } \ \frac{x}{r}[/tex]
DEₓ = k dq x / r³
let's use the Pythagorean theorem to find the distance r
r² = x² + a²
where a is the radius of the ring
we substitute
dEₓ = [tex]k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq[/tex]
we integrate
∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} } ∫ dq
Eₓ = [tex]k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}[/tex]
In the exercise indicate that the electron is very central to the center of the ring
x << a
Eₓ = [tex]k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}[/tex]
if we expand in a series
[tex](\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2[/tex]
we keep the first term if x<<a
Eₓ = [tex]\frac{ k Q}{a^3} \ x[/tex]
the force is
F = q E
F = [tex]- \frac{kQ }{a^3} \ x[/tex]
this is a restoring force proportional to the displacement so the movement is simple harmonic,
F = m a
[tex]- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }[/tex]
[tex]\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x[/tex]
the solution is of type
x = A cos (wt + Ф)
with angular velocity
w² = [tex]\frac{keQ}{m a^3}[/tex]
angular velocity and period are related
w = 2π/ T
we substitute
4π² / T² = \frac{keQ}{m a^3}
T = 2π [tex]\sqrt{\frac{m a^3 }{keQ} }[/tex]
let's calculate
T = 2π [tex]\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }[/tex]
T = 2π pi [tex]\sqrt{320.426 \ 10^{-18} }[/tex]
T = 2π 17.9 10⁻⁹ s
T = 1.12 10⁻⁷ s
a spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
a. 0.0014 meters
b. 0.45 meters
c. 2.2 meters
d. 5.0 meters
show your work
Answer:
I think its B
Explanation:
is it possible to have rainbows during the fine sunny day? Explain your answer.
Answer:
As long as there is rain, a rainbow is possible. Rain is possible on a sunny day, and is known as a sunshower.The rainbow can be observed in a sunny day if the water droplets are present in air and the sun rays pass through it reaches the eye of the observer. In this situation, the observer can see a rainbow.
or
If you happened to look up at the sky this past weekend, you might have noticed a rare and beautiful sight: iridescent rainbow clouds, but not a drop of rain in sight. This phenomenon is known, fittingly, as cloud iridescence or irisation. The effect is not unlike seeing a rainbow painted on the clouds.
The temperature of a body is from 200 to 300C.The change of temp at absolute scale is
Answer:
mark me brainliest
Explanation:
The change of temperature at absolute scale is. A. 3.73 K
Answer:
373K
Explanation:
300°c - 200°c =100°c
Absolute scale means Kelvin scale so
0°c= 273°c
100°c = 100 + 273
=373K
1. Pam has a mass of 48.3 kg and she is at rest on
smooth, level, frictionless ice. Pam straps on
a rocket pack. The rocket supplies a constant
force for 27.3 m and Pam acquires a speed of
62 m/s.
What is the magnitude of the force?
Answer in units of N.
2. What is Pam’s final kinetic energy?
Answer in units of J.
3. A child and sled with a combined mass of 55.7
kg slide down a frictionless hill that is 11.3 m
high at an angle of 29 ◦
from horizontal.
The acceleration of gravity is 9.81 m/s
3. If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s
Answer:
1. F = 3400 N = 3.4 KN
2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3. v = 14.9 m/s
Explanation:
1.
First, we will calculate the acceleration of Pam by using the third equation of motion:
[tex]2as = v_f^2-v_i^2[/tex]
where,
a = acceleration = ?
s = distance = 27.3 m
vf = final speed = 62 m/s
vi = initial speed = 0 m/s
Therefore,
[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]
Now, we will calculate the force by using Newton's Second Law of Motion:
F = ma
F = (48.3 kg)(70.4 m/s²)
F = 3400 N = 3.4 KN
2.
Final kinetic energy is given as:
[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]
[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3.
According to the law of conservation of energy:
[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]
where,
v = speed at bottom = ?
g = acceleration due to gravity = 9.81 m/s²
h = height at top = 11.3 m
Therefore,
[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]
v = 14.9 m/s
Please help... I'm confused on what I represents in terms of solving the total current. Would variable would I be singling out?
Answer:
the researcher say hi for us the best pa the best of us are going out to eat that I can get my money toward a little bit but the best of luck to be at work by then and we will see what the status
9. Cellular respiration occurs in what types of cells?
Answer:
Cellular respiration takes place in the cells of all organisms. It occurs in autotrophs such as plants as well as heterotrophs such as animals. Cellular respiration begins in the cytoplasm of cells. It is completed in mitochondria
Explanation:
Cellular respiration takes place in the cells of all organisms. It happening in autotrophs such as plantas as well as heterotrophs such as animals. Cellular respiration starts in the cytoplasm of cells.
It is finished in mitochondria.
A 0.413 kg block requires 1.09 N
of force to overcome static
friction. What is the coefficient
of static friction?
(No unit)
PLEASE HELP!
Answer:
static friction=0.126
True False: Marke each statement as true or false.
shift
1. Light intensity affects the rate of photosynthesis.
in
2. Energy is required by all organisms for life.
3. The ability of a plant to repair tissue depends on respiration.
4. The gas needed for photosynthesis is carbon dioxide (CO2).
5. Plants only carry on photosynthesis, not respiration.
6. Respiration can occur without photosynthesis.
Answer:
1) true 2) true 3) true 4) true 5) false 6) false
Explanation:
The diagram below shows a person swinging a hammer.
Which position has the least amount of kinetic energy?
A-1
B-2
C-3
D-4
g a mass of 1.3 kg is pushed horizontally against a massless spring with a spring constant of 58 n/m until the spring compresses 19.5 cm if the mass is then released what is the kinetic energy of the mass when it is no longer in contact with the spring ignore friction
Answer: [tex]1.102\ J[/tex]
Explanation:
Given
Mass [tex]m=1.3\ kg[/tex]
Spring constant [tex]k=58\ N/m[/tex]
Compression in the spring [tex]x=19.5\ cm\ or\ 0.195\ m[/tex]
When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.
[tex]\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J[/tex]
The kinetic energy of the mass is 1.102 J.
A chemist measures the flow of charged ions through a circuit. Which of these would increase the current? Select all that apply.
Help me with this please
Answer:
check out of phase
Explanation:
this is my answer
Two parallel slits are illuminated by light composed of two wavelengths, one of which is 657 nm. On a viewing screen, the light whose wavelength is known produces its third dark fringe at the same place where the light whose wavelength is unknown produces its fourth-order bright fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength
Answer:
λ = 5.75 10⁻⁷ mm
Explanation:
This is a slit interference exercise, we analyze each wavelength separately
λ = 657 nm indicate that the third dark pattern
a sin θ = (m + ½) lam
a sin θ = (3 + ½) 657 10⁻⁹
a sin θ = 2299.5 10⁻⁹ nm
for the other wavelength in the same place we have m = 4 bright
a sin θ = m lam
we substitute
2299.5 10⁻⁹ = 4 λ
λ = [tex]\frac{2299.5 \ 10^9 }{ 4}[/tex]
λ = 5.75 10⁻⁷ mm
An electrical insulator is a material that:
A) contains no charge
B) does not allow electrons to flow
C) has more protons than electrons
D) must be a crystal
Answer:
Option B is appropriate for this question
please help, im having trouble can you please explain how to do it
Answer:
independent and the second one with the (-8, 1, -3)
PLEASE HELP I WILL MARK BRAINLYIST!!!!!!!!!
All of the matter in the universe formed after the big bang. Scientists
predicted that hydrogen and helium would be the most abundant elements.
What did scientists discover?
A. The universe has 73% hydrogen and 25% helium, which does not
support the big bang theory
B. Planets are made mostly of other elements, which does not
support the big bang theory
C. Stars are made mostly of other elements, which supports the big
bang theory
D. The universe has 74% hydrogen and 24% helium, which supports
the big bang theory.
Answer:
The answer is D. Here is proof . Plz mark brainliest
Explanation:
Solve the following problem. A force of 5N is being applied to the right, and a force of
3N is being applied to the left. What is the net force?
A- 15 N
B- 8 N
C- 7N
D- 2N
Answer:
D 2N
Explanation:
these are opposite forces so one becomes negative and the other positive that is
-3+5=2 or 5-3=2
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spot on a screen 1.6 m from the slits. Light of wavelength 670 nm is then projected through the same slits. Part A How far from the central bright spot will the second-order maximum of this light be located
Answer:
14.9 mm
Explanation:
We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m
Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m
So, sinθ = mλ/d
Since θ is small, sinθ ≅ tanθ
So,
mλ/d = L/D
d = mλD/L
Substituting the values of the variables into the equation, we have
d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m
d = 2448 × 10⁻⁹ m²/0.017 m
d = 144000 × 10⁻⁹ m
d = 1.44 × 10⁻⁴ m
d = 0.144 × 10⁻³ m
d = 0.144 mm
Now, for the second-order maximum, m' of the 670 nm wavelength of light,
m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m
So, L' = m'λ'D/d
So, substituting the values of the variables into the equation, we have
L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m
L' = 2144 × 10⁻⁹ m²/0.144 × 10⁻³ m
L' = 14888.89 × 10⁻⁶ m
L' = 0.01488 m
L' ≅ 0.0149 m
L' = 14.9 mm