Answer:
No, the acceleration is not always zero.
Explanation:
It does not mean that the acceleration of the particle is zero.
The velocity of wave is different from the velocity of particle.
The acceleration of wave is different from the acceleration of particle.
the acceleration of the particle is given by
[tex]a =- w^2 y[/tex]
where, w is the angular frequency and y is the displacement from the mean position.
So, the acceleration is zero at mean position only and it varies as the position changes.
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.
Answer:
the distance from the cornea where a very distant object will be imaged is 23.35 mm
Explanation:
Given the data in the question;
For a spherical refracting surface;
[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium
[tex]d_0[/tex] is the object distance
[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium
[tex]d_i[/tex] is the image distance
R is the radius of curvature
Now, let [tex]d_0[/tex] = ∞, such that;
[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
we make [tex]d_i[/tex] subject of the formula
[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )
[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )
given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00
so we substitute
[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )
[tex]d_i[/tex] = 9.165 / 0.41
[tex]d_i[/tex] = 23.35 mm
Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm
Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.
Answer:
An object in either state of equilibrium must have no net force acting on it.
Explanation:
Answer: An object in either state of equilibrium must have no net force acting on it.
Explanation:
What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm
Answer:
The submarine's maximum safe depth in sea water is 801.678 m.
Explanation:
P=Po+(rho)*g*h
Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)
Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2
Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa
Initial Pressure= 1atm= 101.3kPa= 101300Pa
Water Density (rho) = 1000kg/m^3
Gravity= 9.8m/s^2
So rearranging for h= (P-Po)/((rho)*g)
h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m
15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?
Answer: (15 - 0)/1.8 = 8. 33m/s^2
Explanation:
The acceleration of the racehorse is 8.33 m/s²
The given parameters;
initial velocity of the racehorse, u = 0
final velocity of the racehorse, v = 15 m/s
time of motion of the horse, t = 1.8 s
The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]
Thus, the acceleration of the racehorse is 8.33 m/s²
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3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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Which of the following accurately describes circuits?
options:
A)
In a parallel circuit, there's only one path for the current to travel.
B)
In a series circuit, the amount of current passing through each part of the circuit may vary.
C)
In a series circuit, the current can flow through only one path from start to finish.
D)
In a parallel circuit, the same amount of current flows through each part of the circuit.
' C ' is the only correct statement.
Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
(a+b)[tex]x^{2}[/tex]
In which situation should a parent be proactive and act to assume responsibility?
Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.
Explanation: i picked this and it is correct, you’re welcome:)
Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave
Ez Physics question will mark brainliest.
Answer:
The answer is B. longitude wave
Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is
Explanation:
Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]
[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]
[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]
The work function [tex]\phi[/tex] is then
[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]
[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]
The work function of the surface is equal to 3.98 × 10⁻¹⁹J.
What are frequency and wavelength?The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.
The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.
The expression between wavelength (λ), frequency, and speed of light (c) is:
c = νλ
Given, the wavelength of the light, ν = 500 nm
The frequency of the light can determine from the above-mentioned relationship:
ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz
The work function = h ν = 6 × 10¹⁴ × 6.626 × 10⁻³⁴
φ = 3.98 × 10⁻¹⁹J
Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.
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A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.
Answer:
I think D) less than 50 years
Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.
Who is an astronaut?An astronaut observes and performs the experiments based on the universe.
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.
Due to special relativity, between space and Earth, both moving with different speeds.
The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.
So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.
Thus, the correct option is d.
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A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
a baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn inclined at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline
Explanation:
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g .
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices
Answer:
check photo for solve
Explanation:
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?
Answer:
The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.
Explanation:
Consider newtonian mechanics here.
Dynamic equation is
The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.
we'll consider newtonian mechanics here.
so the dynamic equations is S = ut + 0.5at^2
we know u=0.8
S=1.2+0.9
t=1.2/1.5
find a.
Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction
Answer:
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Explanation:
The expression for the angle of refraction is
n₁ sin θ₁ = n₂ sin θ₂
refractive index n₁ is for incident light and n₂ is for transmitted light.
We have two cases
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
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Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1
Answer:
x= 2*I1*d/(I1+I2) meter
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.
Answer:
The distance between mirror and you is 2 ft.
Explanation:
diameter, d = 8 ft
radius of curvature, R = 4 ft
magnification, m = 0.5
focal length, f = R/2 = 4/2 = 2 ft
let the distance of object is u and the distance of image is v.
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]
Use the formula of magnification
[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]
State the law of conservation of momentum
Explanation:
Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
[tex]v=(6ti+6k)\ m/s[/tex]
Explanation:
Given that,
The position of a particle is given by :
[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]
Let us assume we need to find its velocity.
We know that,
[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]
So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].
A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event
Answer:
The rise in temperature is 0.06 K.
Explanation:
mass of bullet, m = 15 g
initial speed, u = 865 m/s
final speed, v = 534 m/s
mass of water, M = 13.5 kg
specific heat of water, c = 4200 J/kg K
The change in kinetic energy
[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]
According to the conservation of energy, the change in kinetic energy is used to heat the water.
K = m c T
where, T is the rise in temperature.
3473 = 13.5 x 4200 x T
T = 0.06 K
a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection
Answer:
Answer:30 cm
Answer:30 cmExplanation:
Answer:30 cmExplanation:Given=ROC= 60cm
Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed
Answer:
c.
magnitude of displacement
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
