Explanation:
Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles .
A student trying to calculate the coefficient of friction between a box and a surface. She measures that the 80kg box will slide if the student pushes with a force greater than 400n.
Answer:
[tex]\mu_s=0.51[/tex]
Explanation:
Coefficient of friction
The static coefficient of friction is a measure of the force needed to start moving an object from rest along a rough surface.
It's calculated as:
[tex]\displaystyle \mu_s=\frac{F_r}{N}[/tex]
Where Fr is the friction force and N is the normal force exerted by the surface over the object.
In the absence of any other vertical force, the normal is equal to the weight of the object:
[tex]N = W = m.g[/tex]
The box has a mass of m=80 Kg, thus the normal force is:
[tex]N = 80\ Kg * 9.8\ m/s^2[/tex]
N = 784 N
The student needs to push with a force of 400 N to make the box move. That is the force that barely outcomes the friction force. Thus:
[tex]F_r=400\ N[/tex]
Calculating the coefficient:
[tex]\displaystyle \mu_s=\frac{400}{784}[/tex]
[tex]\mathbf{\mu_s=0.51}[/tex]
Which symbol and unit of measurement are used for the electric current?
symbol: A; unit: I
symbol: C; unit: A
symbol: I; unit: C
symbol: I; unit: A
Answer:
Symbol: I unit: A
Explanation:
the formula for current is I = v/r
I: Current
V: voltage
R; Resistance
Symbol and unit of measurement are used for the electric current are symbol: I; unit: A
What is electric current ?"Electric Current is the rate of flow of electrons in a conductor. The SI Unit of electric current is the Ampere. Electrons are minute particles that exist within the molecular structure of a substance. Sometimes, these electrons are tightly held, and the other times they are loosely held."
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A 150.0 kg cart rides down a set of tracks on four solid steel wheels, each with radius 20.0 cm and mass 45.0 kg. The tracks slope downward at an angle of 17 ∘ to the horizontal. If the cart is released from rest a distance of 27.0 m from the bottom of the track (measured along the slope), how fast will it be moving when it reaches the bottom
Answer:
13.4 m/sExplanation:
given
Mass of cart= 150kg
mass of each wheel=45kg
mass of 4 wheels= 180kg
angle of the track= 17 ∘
distance of track= 27m
The height of the tracl is calculated thus:
sin 17° = h / 27
h = sin 17*27
h=7.89m
"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "
1. Potential energy at top= (M+4m)gh
2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²
3. rotational energy of wheels at bottom= 4(1/2 Iω²)
The total is expressed as
(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) -------------1
we know that I = mr² / 2
Put I= mr² / 2
(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)
(M+4m)gh = 1/2 (M+4m)v² + m r² ω²
we know that v²= r² ω²
(M+4m)gh = 1/2 (M+4m)v² + m v²
(M+4m)gh = v² (M/2 + 2m + m)
(M+4m)gh = v² (M/2 + 3m)
v = √[(M+4m)gh / (3m + M/2)]
v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]
v=√25516.26/142.5
v=√179.06
v = 13.4 m/s
90% of cancer originate from
Carcinoma refers to a malignant neoplasm of epithelial origin or cancer of the internal or external lining of the body. Carcinomas, malignancies of epithelial tissue, account for 80 to 90 percent of all cancer cases. Epithelial tissue is found throughout the body.
A force of 35.0 N is required to start a 6.0-kg box movingacross a horizontal concrete floor. (a) What is the coefficientof static friction between the box and the floor?(b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2 . What is the coefficient of kinetic friction.
Answer:
Explanation:
Frictional force acting on the box = 35 N
If μ₁ be the coefficient of static friction and μ₂ be the kinetic friction
μ₁ mg = 35
μ₁ = 35 / 6 x 9.8
= .6
b )
net force acting on box to produce acceleration of .6 m /s²
= 6 x .6 = 3.6 N
net force acting on the box when it is accelerating
= 35 - μ₂ mg
35 - μ₂ x 6 x 9.8 = 3.6
35 - μ₂ x 58.8 = 3.6
μ₂ = .53
A particular radiograph was produced using 6 mAs and 110 kVp with an 8:1 ratio grid. The radiograph is to be repeated using a 16:1 ratio grid. What should be the new mAs?
(A) 3 mAs
(B) 6 mAs
(C) 9 mAs
(D) 12 mAs
Answer:
(C) 9 mAs
Explanation:
The computation of the new mAs is shown below:
Let us assume the New mAs be x
For adjusting the factors of exposure basically we compared the old one with the new one i.e. presented below:
6 ÷ x = 4 ÷ 6
where
6 = old mAs
4 = old grid factor
x = new mas
6 = new grid factor
Now solve this above equation with the help of cross multiplication
4x = 36
x = 9 mAs
Here we used the 16:1 ratio grid
Hence, the correct option is C. 9 mAs
Referring to the sketch of a planet around the sun, Area A is three times that of Area B. Compare the times required for the planet to travel from Point 1 to Point 2 and from Point 3 to Point 4 and select the letter of the correct answer.
Answer:
tA is three times tB
Explanation:
tA is three times tB
Referring to the sketch of a planet around the sun, Area A is three times that of Area B. Compare the times required for the planet to travel from Point 1 to Point 2 and from Point 3 to Point 4 and select the letter of the correct answer.
For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
A) Find the energy in electron volts for a particle with this wavelength if the particle is a photon.
B) Find the energy in electron volts for a particle with this wavelength if the particle is an electron.
C) Find the energy in electron volts for a particle with this wavelength if the particle is an alpha particle (m=6.64×10−27kg).
Answer:
A) E = 4.96 x 10³ eV
B) E = 4.19 x 10⁴ eV
C) E = 3.73 x 10⁹ eV
Explanation:
A)
For photon energy is given as:
[tex]E = hv[/tex]
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = energy of photon = ?
h = 6.625 x 10⁻³⁴ J.s
λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Therefore,
[tex]E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}[/tex]
[tex]E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})[/tex]
E = 4.96 x 10³ eV
B)
The energy of a particle at rest is given as:
[tex]E = m_{0}c^2[/tex]
where,
E = Energy of electron = ?
m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
[tex]E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\[/tex]
[tex]E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]
E = 4.19 x 10⁴ eV
C)
The energy of a particle at rest is given as:
[tex]E = m_{0}c^2[/tex]
where,
E = Energy of alpha particle = ?
m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
[tex]E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\[/tex]
[tex]E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]
E = 3.73 x 10⁹ eV
A) The energy in electron volts for a particle with this wavelength if the particle is a photon is; .E = 4969.5 eV or 4.9695 keV
B) The energy in electron volts for a particle with this wavelength if the particle is an electron is; E = 23.58 eV
C) E = 0.003306 eV
A) The formula for the energy here is;
E = hc/λ
where;
h is planck's constant = 6.626 × 10⁻³⁴ J.s
c is speed of light = 3 × 10⁸ m/s
λ is wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Thus;
E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(0.25 x 10⁻⁹)
79.512 × 10⁻¹⁷ J
converting to eV gives;
E = (79.512 × 10⁻¹⁷)/(1.6 × 10⁻¹⁹)
E = 4969.5 eV or 4.9695 keV
B) Formula for the energy if the particle is an electron is;
E = h²/(2mλ²)
where m = 9.31 × 10⁻³¹ kg
E = (6.626 × 10⁻³⁴)²/(2 × 9.31 × 10⁻³¹ × (0.25 x 10⁻⁹)²)
E = 37.726 × 10⁻¹⁹ J
Converting to eV gives;
E = (37.726 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)
E = 23.58 eV
C) Mass of alpha particle is; m = 6.64 × 10⁻²⁷ kg
E = h²/(2mλ²)
where m = 6.64 × 10⁻²⁷ kg
E = (6.626 × 10⁻³⁴)²/(2 × 6.64 × 10⁻²⁷ × (0.25 x 10⁻⁹)²)
E = 52.896 × 10⁻²³ J
Converting to eV gives;
E = (52.896 × 10⁻²³)/(1.6 × 10⁻¹⁹)
E = 0.003306 eV
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The aurora borealis is caused by the ____.
A. mesosphere
B. stratosphere
c. thermosphere
D. troposphere
Answer: ionosphere
Explanation: aurora boreal is is caused by the ionosphere which is a part of the thermosphere
A 5.0 kg hammer strikes a 0.25 kg nail with a force of 10.0 N causing the nail to accelerate at 40.0 m/s^2. What is the acceleration of the hammar?
Answer:
2 m/s².
Explanation:
The following data were obtained from the question:
Mass (m) of hammer = 5 Kg
Mass (m) of nail = 0.25 kg
Force (F) applied = 10 N
Acceleration (a) of nail = 40 m/s².
Acceleration (a) of hammer =?
From Newton's third law which states that to every action, there is an equal but opposite reaction. This implies that the force applied by the hammer on the nail is exactly the force applied by the nail on the hammer. Thus, we can obtain the acceleration of the hammer as follow:
Mass (m) of hammer = 5 Kg
Force (F) applied = 10 N
Acceleration (a) of hammer =?
F = ma
10 = 5 × a
Divide both side by 5
a = 10/5
a = 2 m/s²
Thus, the acceleration of the hammer is 2 m/s².
1) What do (x) and (y) symbolize?
a) What equations belong to each one?
b) How does horizontal motion effect vertical motion and vis versa?
c) What do both horizontal and vertical motion use?
Answer:
x is vertical and y is horizontal
Explanation:
mention three features of the Constitution and a note about them
Answer:
Explanation:(1) a constitution is a supreme law of the land, (2) a constitution is a framework for government; (3) a constitution is a legitimate way to grant and limit powers of government officials. Constitutional law is distinguished from statutory law.
A 0.208 kg particle with an initial velocity of 1.26 m/s is accelerated by a constant force of 0.766 N over a distance of 0.195 m. Use the concept of energy to determine the final velocity of the particle. (It is useful to double-check your answer by also solving the problem using Newton's Laws and the kinematic equations.)
Answer:
Explanation:
Initial kinetic energy of particle
= 1/2 m V²
= .5 x .208 x 1.26²
= .165 J
Work done by force = force x displacement
= .766 x .195
= .149 J
This energy will be added up .
Total final kinetic energy
= initial kinetic energy + work done on the particle
= .165 + .149 J
= .314 J .
How does the moon cause tides?
Answer:
High and low tides are caused by the Moon. The Moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the Moon and the side farthest from the Moon. These bulges of water are high tides.
Explanation:
A 5 kg rock is dropped down a vertical mine shaft. How long does it take to reach the bottom, 79 meters below?
Answer:
The time for the rock to reach the bottom is 4.02 seconds.
Explanation:
Given;
mass of the rock, m = 5 kg
height of the rock fall, h = 79 meters
The time to drop to the given height is given by;
[tex]t = \sqrt{\frac{2h}{g} }[/tex]
where;
t is the time to fall to the bottom
g is acceleration due to gravity = 9.8 m/s²
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s[/tex]
Therefore, the time for the rock to reach the bottom is 4.02 seconds.
If Batman jumps off a 50 m tall building at 30m/s to save a civilian...
a) How long does it take him to reach the civilian?
b) How far from the bottom of the building does he go?
Answer:
Use
[tex]t = \: \sqrt{2h \div g } [/tex]
[tex]x = u \times \sqrt{2h \div g} [/tex]
if an object travels at constant speed in a circular path the acceleration of the object is
Answer:
centripetal and of constant magnitude
Explanation:
The acceleration of an object traveling on a circular path at constant speed is directed towards the center of the circle (centripetal) and of constant magnitude equal to the square of the object's speed divided by the radius of the circle.
Question 4
Why do some competitive swimmers shave their heads and bodies?
А
To decrease weight
B.
to increaase blood flow
C
to decrease friction
D
to increase buoyancy
Answer:
C
Explanation:
A 4.00-kg mass and a 9.00-kg mass are being held at rest against a compressed spring on a frictionless surface. When the masses are released, the 4.00-kg mass moves to the east with a speed of 2.00 m/s. What is the velocity of the 9.00-kg mass after the masses are released?
A) 0.500 m/s east
B) 0.500 m/s west
C) 4.50 m/s west
D) 0.888 m/s west
E) 2.00 m/s west
Answer:
B) 0.500 m/s West
brainlist answer
Three beakers are of identical shape and size , one beaker is pained matt black, one is dull white and one is gloss white . The beakers are filled with boiling waterin which beaker does the water cool more quickly? Give reason. ( i really need the answer quickly)
Answer:
The beaker that is matt black heats more quickly because black attracts more heat.
Explanation:
Three beakers are of identical shape and size, one beaker is pained matt black, one is dull white and one is gloss white . The beakers are filled with boiling water. The water in the matt black beaker cools more quickly.
The rate of cooling of an object is influenced by its surface properties and color. In this case, the matt black beaker will cool the water more quickly compared to the dull white and gloss white beakers.
The reason behind this lies in the concept of thermal radiation. Darker colors, such as matt black, absorb and emit thermal radiation more efficiently than lighter colors. When the boiling water is placed in the matt black beaker, the black surface absorbs a greater amount of thermal radiation from the water and its surroundings. This increased absorption accelerates the transfer of heat energy from the water to the beaker, leading to faster cooling.
On the other hand, the dull white and gloss white beakers reflect more thermal radiation, absorbing and emitting less heat. As a result, the water in these beakers cools at a slower rate compared to the matt black beaker.
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An object weighs 3 lb. at 10 earth radii from its center. What is the object's weight on the earth's surface?
______ lb.
300
30
3
3,000
Answer:
3
Explanation:
For an investigation a student records data about four unknown substances.
data, for, unknown, substances, substance, mass, grams, volume, centimeters, cubed, density, grams, per, centimeter, cubed, 1, 6.95, 4.0, 2, 4.54, 2.0, 3, 5.40, 3.0, 4, 10.35, 5.0,
The student then calculates the densities of the unknown substances and compares them with the table of densities of known substances shown below.
densities, of, some, known, substances, substance, density, grams, per, centimeter, cubed, calcium, 1.54, carbon, 2.27, magnesium, 1.74, phosphorus, 1.82, platinum, 21.46, sulfur, 2.07,
Which unknown substance is most likely carbon?
Answer:
omparing the values of the strung density, the one that is closest to carbon is number 2
Explanation:
In this exercise we are given the mass and volume of a body, we are asked to calculate the density, using the equation
ρ = m / V
in the third and fourth column of the table is the density and the substance with the closest value
# mass volume density material
(gr) (cm³) (gr/cm³)
1 6.95 4.0 1.74 magnesium
2 4.54 2.0 2.27 carbon
3 5.40 3.0 1.80 phosphorus
4 10.35 5.0 2.07 sulfur
When comparing the values of the strung density, the one that is closest to carbon is number 2
Two satellites are in orbit around a planet. One satellite has an orbital radius of 8.0×1 0 6 m. The period of revolution for this satellite is 1.0×1 0 6 s. The other satellite has an orbital radius of 2.0×1 0 7 m. What is this satellite’s period of revolution?
This question involves the concept of the orbital period.
The period of revolution of the second satellite is "3.95 x 10⁶ s".
ORBITAL PERIODFirst, we will consider the orbital period of the first satellite:
[tex]T_1=\sqrt{\frac{4\pi^2 r_1^3}{GM}}[/tex]
where,
T₁ = orbital period of frirst satellite = 1 x 10⁶ sr₁ = orbital radius for first satellite = 8 x 10⁶ mM = mass of planet = ?G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²Therefore,
[tex]1\ x\ 10^6\ s=\sqrt{\frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(M)}}\\\\M = \frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1\ x\ 10^6\ s)^2}\\\\M = 3.03\ x\ 10^{20}\ kg[/tex]
Now, we find out the orbital period of the second satellite:
[tex]T_2=\sqrt{\frac{4\pi^2 r_2^3}{GM}}[/tex]
where,
T₂ = orbital period of second satellite = ?r₁ = orbital radius for second satellite = 2 x 10⁷ mTherefore,
[tex]T_2=\sqrt{\frac{4\pi^2(2\ x\ 10^7\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3.03\ x\ 10^{20}\ kg)}}[/tex]
T₂ = 3.95 x 10⁶ s
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Which of the following represents a chemical change when using bread in a meal?
F. Removing the crust of the bread when making a sandwich.
G. Placing the bread in a toaster to make toast and applying butter on it afterwards
H. Cutting the bread in half to make two sandwiches
J. Placing mayonnaise, ketchup, and mustard on the bread before the meat
Answer:Well toasting it makes it a solid so that’s a chemical change that happens! Hope this helped!
1. An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s. How fast will he be moving backward just after releasing the ball?2. Which of the following best describes why you can analyze this event using conservation of momentum?A. The throwing action is quick enough that external forces may be ignored.B. External forces don't act on the system during the jump.C. Conservation of momentum is always the best way to analyze motion.
Answer:
1) 0.0806 m/s
2) External forces don't act on the system during the jump
Explanation:
velocity of ball ( Vbf )= 15 m/s
mass of quarter back ( Mq )= 80 kg
mass of football ( Mb ) = 0.43 kg
A) How fast will be be moving backward just after releasing the ball
we can determine this speed with the use of conservation of momentum equation
= Mb ( Vbf - Vbi ) = Mq ( Vq )
where Vq = Mb ( Vbf - Vbi ) / Mq
= 0.43 ( 15 m/s - 0 ) / 80 kg
= 0.0806 m/s
B) External forces don't act on the system during the jump
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it.
Part 1) What should be the focal length of this lens?
Answer in units of cm.
Part 2) What is the power of the needed corrective lens?Answer in units of diopters.
Answer:
a) 0.3 m
b) 3.3 diopters
Explanation:
Given that
Object distance is 26cm from the eye.
Image distance is 105 cm
From the above, we will use the formula
1/f = 1/v + 1/u
where
f is the focal length of the lens,
v is the image distance,
u is the object distance.
Since the image is a virtual image, we will give our v a negative sign. So, on calculating, we have
1/f = 1/25 - 1/151
1/f = 0.0333
f = 1/0.0333
f = 30.03 cm
f = 0.3 m
b) The power of the needed corrective lens is the reciprocal of the focal length in metres;
1/0.3 = 3.3 diopters
This is usually in diopters
What did Rutherford conclude from the motion of the particles when shot through a thin gold foil
Answer:
He concluded the fact that most alpha particles went straight through the foil is evidence for the atom being mostly empty space. A small number of alpha particles being deflected at large angles suggested that there is a concentration of positive charge in the atom.
Explanation:
6) There are two reactants in a chemical equation, and one product. The mass of the product is 40g. The mass of the first reactant is 16g. What must the mass of the second reactant be, if the equation is to follow the law of conservation of mass?
Answer:
24g
Explanation:
mass of the product is 40g.
The mass of the first reactant is 16g
The mass of the second reactant = ?
Let the first reactant = A
Let the second reactant = B
Let the product = C
Equation of reaction;
A + B → C
16g 40g
The mass of B must be = 40 -16 = 24g
The acceleration due to gravity on Jupiter is 23.1 m/s2, which is about twice the acceleration due to gravity on Neptune.
Which statement accurately compares the weight of an object on these two planets?
An object weighs about one-fourth as much on Jupiter as on Neptune.
An object weighs about one-half as much on Jupiter as on Neptune.
An object weighs about two times as much on Jupiter as on Neptune.
An object weighs about four times as much on Jupiter as on Neptune.
Answer:
I believe its C
Explanation:
Answer:
C-An object weighs about two times as much on Jupiter as on Neptune.
Explanation:
on edg
A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?
Answer:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
Explanation:
We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:
[tex]y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (1)
Where:
[tex]y(t)[/tex] - Position of the mass as a function of time, measured in meters.
[tex]A[/tex] - Amplitude, measured in meters.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]t[/tex] - Time, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
The spring is now calculated by Hooke's Law, that is:
[tex]k = \frac{m\cdot g}{\Delta y}[/tex] (2)
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta y[/tex] - Deformation of the spring due to gravity, measured in meters.
If we know that [tex]m=1.65\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta y = 0.260\,m[/tex], then the spring constant is:
[tex]k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}[/tex]
[tex]k = 62.237\,\frac{N}{m}[/tex]
If we know that [tex]A = 0.130\,m[/tex], [tex]k = 62.237\,\frac{N}{m}[/tex], [tex]m=1.65\,kg[/tex], [tex]x(t) = 0\,m[/tex] and [tex]\phi = 0\,rad[/tex], then (1) is reduced into this form:
[tex]0.130\cdot \cos (6.142\cdot t)=0[/tex] (1)
And now we solve for [tex]t[/tex]. Given that cosine is a periodic function, we are only interested in the least value of [tex]t[/tex] such that mass reaches equilibrium position. Then:
[tex]\cos (6.142\cdot t) = 0[/tex]
[tex]6.142\cdot t = \cos^{-1} 0[/tex]
[tex]t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s[/tex]
[tex]t \approx 0.255\,s[/tex]
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
The time taken for the spring to reach new equilibrium position is 0.26 s.
The given parameters;
mass, m = 1.65 kgextension of the string, x = 0.26 mdisplacement with time x(t) = 0.13The general wave equation is given as;
[tex]y(t) = A\ cos(\omega t \ + \phi)[/tex]
The angular frequency is given as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\[/tex]
The spring constant is calculated as;
[tex]F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{1.65 \times 9.8}{0.26} \\\\k = 62.2 \ N/m[/tex]
The angular frequency is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{62.2}{1.65} }\\\\\omega = 6.14 \ rad/s[/tex]
The time taken for the spring to reach new equilibrium position is calculated as follows;
[tex]y(t) = A \ cos(\omega t)\\\\0 = 0.13 \times cos(6.14t)\\\\6.14t = cos^{-1}(0)\\\\6.14t = 1.57 \ rad\\\\t = \frac{1.57 \ rad}{6.14 \ rad/s} \\\\t = 0.26 \ s[/tex]
Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.
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