Answer:
b
Explanation:
A thermodynamic-ally favored reactions are nothing but spontaneous reactions. In spontaneous reactions ΔH (enthalpy) is always negative and ΔS( entropy) is always positive.
Now all thermodynamic-ally favored reactions does not occur rapidly because the reactants still need to break bonds and overcome the activation energy in order to form products.
Thermodynamically favored reactions do not necessarily occur rapidly due to the need to overcome activation energy, which is a kinetic factor unrelated to the thermodynamics of the reaction. The rate of reaction is influenced by how frequently reactants can attain the transition state, which may be facilitated by catalysts and temperature.
Not all thermodynamically favored reactions occur rapidly because of kinetics related to activation energy. While thermodynamics can indicate whether a reaction is favorable or not, it does not take into account the energy required to reach the transition state where reactants are converted into products.
Option b is correct because reactants need to overcome the activation energy barrier and break bonds to form products. The speed at which a reaction approaches its equilibrium does not depend on the reaction's equilibrium constant but rather on the energy profile of the reaction pathway and the frequency at which reactants attain the critical transition state.
Catalysts can lower the activation energy, hence increasing the rate of a reaction without altering the thermodynamic favorability. Conditions such as temperature also play a role since molecules at higher temperatures have more kinetic energy, which increases the probability of achieving the energy threshold needed to initiate the reaction.
Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A)-0 B)-370JC) -190 J D-1800 J
Answer:
the network done by friction on a box that moves in a complete circle is 185.7 joules
Explanation:
Step one
Given
Radius of circle =1.82m
Circumference of the circle =2*pi*r
=2*3.142*1.82=11.43
Hence distance =11.43m
Coefficient of friction u=0.25
Weight of box =65N
We know that work =force*distance
But the limiting force =u*weight
Hence the net work done by friction
Wd=0.25*65*11.43
Wd=185.7 joules
The work done by friction on a box moving in a complete circle is zero because the force of friction is always perpendicular to the direction of the box’s displacement.
Explanation:The subject of this question is the work done by friction on a box moving in a circle. When an object moves in a circle, it is the force of friction that prevents the object from sliding off the circular path and helps maintain a curved path. However, it is important to note that friction does no work when an object moves in a complete circle because the force of friction is at every point perpendicular to the direction of the box’s velocity.
To calculate work, we generally use the formula W = F * d * cos(θ), where F is the force (which is the kinetic friction force in this case), d is the displacement of the box (the path covered), and θ is the angle between the force and the direction of displacement. As the box moves in a complete circle, the angle is always 90 degrees. The cosine of 90 degrees is 0, hence the work done by kinetic friction in this case would be zero (0).
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Air conditioners are rated by their coefficient of performance at 61 ∘ F inside temperature and 99 ∘ F outside temperature. An efficient but realistic air conditioner has a coefficient of performance of 3.2. What is the maximum possible coefficient performance?
Answer:
the maximum possible coefficient performance is 13.7
Explanation:
inside temperature, [tex]T_{C}[/tex] = 61 F = 289.26 K
outside temperature, [tex]T_{H}[/tex] = 99 F = 310.37 K
coefficient of performance, COP (real) = 3.2
according to Carnot's theorem, the coefficient of performance is
[tex]COP_{max}[/tex] = [tex]\frac{T_{C} }{T_{H}-x_{C} }[/tex]
where
[tex]T_{C}[/tex] is cold temperature
[tex]T_{H}[/tex] is hot temperature
thus,
[tex]COP_{max}[/tex] = [tex]\frac{289.26}{310.37-289.26}[/tex]
= 13.7
A ball (mass 0.40 kg) is initially moving to the left at 30 m/s. After hitting the wall, the ball is moving to the right at 20 m/s. What is the impulse of the net force on the ball during its collision with the wall?
Answer:
Impulse, J = 20 m/s
Explanation:
Given that,
Mass of the ball, m = 0.4 kg
Initial speed of the ball, u = -30 m/s (left)
Final speed of the ball after hitting, v = 20 m/s (right)
Let J is the impulse of the net force on the ball during its collision with the wall. The change in momentum of an object is equal to the impulse imparted to it. It is given by :
[tex]J=m(v-u)[/tex]
[tex]J=0.4\ kg(-30-20)\ m/s[/tex]
J = -20 m/s
So, the magnitude of impulse of the net force on the ball during its collision with the wall is 20 m/s.
The impulse on the ball is 20 Ns.
What is impulse?This can be defined as the product of force and time. The impulse of a force acting on a body is also equal to the change in momentum of the body
To calculate the impulse of the net force on the ball during the collision, we use the formula below.
Formula:
I = m(v-u)............ Equation 1Where:
I = Impulse on the ballv = Final velocity of the ballu = Initial velocity of the ballm = mass of the ball.From the question,
Given:
m = 0.40 kgv = 20 m/s u = -30 m/s (to the right)Substitute these values into equation 1
I = 0.40[20-(-30)]I = 0.40(20+30)I = 0.40(50)I = 20 Ns.Hence, the impulse on the ball is 20 Ns.
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While a car is stopped at a traffic light in a storm, raindrops strike the roof of the car. The area of the roof is 5.0 m2. Each raindrop has a mass of 3.7 ✕ 10−4 kg and speed of 2.5 m/s before impact and is at rest after the impact. If, on average at a given time, 150 raindrops strike each square meter, what is the impulse of the rain striking the car?
Answer:
J = 0.693 N.s
Explanation:
The impulse of one single drop is given by:
J1 = m*(Vf - Vo) where Vf = 0
[tex]J1 = -9.25*10^{-4}N.s[/tex]
The magnitude of the total impulse will be:
Jt = J1 * 150 * 5
Jt = 0.693 N.s
A child bounces in a harness suspended from a door frame by three elastic bands. If each elastic band stretches 0.300 m while supporting a 7.15-kg child at rest, what is the force constant for each elastic band? Assume that each spring supports 1/3 of the child's weight.
Answer:
The force constant for each elastic band is 77.93 N/m
Explanation:
Hooke's law of a spring or an elastic band gives the relation between elastic force (Fe) and stretching (x), the magnitude of that force is:
[tex] F_{e}= kx [/tex] (1)
With k, the elastic force constant. The three elastic bands support the child’s weight (W) and maintain him at rest, so by Newton’s second law for one of the elastic bands:
[tex] \sum F=0 [/tex] (2)
[tex]\frac{W}{3}-F_{e}=0\Rightarrow F_{e}=\frac{W}{3} [/tex] (3)
Using (1) on (3):
[tex] kx=\frac{W}{3}\Rightarrow k=\frac{mg}{3x} [/tex](4)
[tex] k=\frac{7.15\,kg*9.81\,\frac{m}{s^{2}}}{3*0.300\,m}\simeq\mathbf{77.93\frac{N}{m}} [/tex]
The force constant of each elastic band is 77.87 N/m.
To find the force constant for each elastic band, we use Hooke's law
Hooke's lawHooke's law states that the force applied to an elastic material is directly proportional to its extension provided the elastic limit is not exceeded
W = ke............ Equation 1Where:
W = Weight of the child on one elastic bandk = Force constante = extensionMake k the subject of the equation
k = W/e.............. Equation 2From the question,
Given:
W = (7.15×9.8)/3 = 23.36 Ne = 0.3 mSubstitute these values into equation 2
k = 23.36/0.3k = 77.87 N/mHence, The force constant of each elastic band is 77.87 N/m.
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The shear modulus of steel is 8.1 × 1010 N/m2. A steel nail of radius 7.5 × 10–4 m projects 0.040 m horizontally outward from a wall. A man hangs a wet raincoat of weight 28.5 N from the end of the nail. Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?
Answer:
[tex]\Delta\ y =7.96 \times 10^{-6}\ m[/tex]
Explanation:
given,
shear modulus of steel = 8.1 × 10¹⁰ N/m²
radius of steel nail = 7.5 × 10⁻⁴ m
Projection outward = 0.040 m
Weight of wet raincoat = 28.5 N
Vertical deflection of other end of the nail = ?
we know
[tex]G = \dfrac{\dfrac{F}{A}}{\dfrac{\Delta y}{L}}[/tex]
[tex]\Delta\ y = \dfrac{FL}{AG}[/tex]
A = π r²
A = π x (7.5 x 10⁻⁴)²
A = 1.767 x 10⁻⁶ m²
[tex]\Delta\ y =\dfrac{28.5 \times 0.04}{1.767\times 10^{-6}\times 8.1\times 10^{10}}[/tex]
[tex]\Delta\ y =7.96 \times 10^{-6}\ m[/tex]
Thus, the vertical deflection of the other end of the nail is[tex]\Delta\ y =7.96 \times 10^{-6}\ m[/tex]
The shear modulus of steel, and applying the necessary formulas, we calculated the vertical deflection due to the weight of the raincoat. This means that the vertical deflection of the other end of the nail is approximately 7.968 μm.
To find the vertical deflection of the steel nail when a weight is hung from it, we can use the relationship between shear deformation, shear modulus, and applied force. The steps to solve the problem are as follows:
Step 1: Identify the given values.
Shear modulus of steel (G) = 8.1 × 10¹⁰ N/m² Radius of the nail (r) = 7.5 × 10⁻⁴ m Length of the nail projecting from the wall (L) = 0.040 m Weight of the raincoat (F) = 28.5 NStep 2: Calculate the cross-sectional area (A) of the nail.
The area can be computed using the formula for the area of a circle:
[tex]A = rac{ ext{π} imes r^2}{1}[/tex]
Substituting the value for radius:
[tex]A = ext{π} imes (7.5 × 10^{-4} ext{ m})^2[/tex]
[tex]A \approx 1.7671 × 10^{-6} ext{ m}^2[/tex]
Step 3: Calculate the shear stress (τ) in the nail.
Shear stress is defined as the force applied per unit area. Thus:
[tex]τ = rac{F}{A}[/tex]
Substituting the force and area:
[tex]τ = rac{28.5 ext{ N}}{1.7671 × 10^{-6} ext{ m}^2}[/tex]
[tex]τ \approx 1.6168 × 10^{7} ext{ N/m}^2[/tex]
Step 4: Calculate the angle of deflection (θ) in radians.
The shear strain (γ) is related to shear stress (τ) via the shear modulus (G):
[tex]γ = rac{τ}{G}[/tex]
Substituting in the values:
[tex]γ = rac{1.6168 × 10^{7} ext{ N/m}^2}{8.1 × 10^{10} ext{ N/m}^2}[/tex]
[tex]γ \approx 0.0001992[/tex]
Step 5: Calculate the vertical deflection (δ) at the end of the nail.
Deflection under shear can be computed using the formula:
[tex]δ = γ imes L[/tex]
So substituting in our calculated shear strain and the length:
[tex]δ = 0.0001992 imes 0.040 ext{ m}[/tex]
[tex]δ \approx 7.968 × 10^{-6} ext{ m}[/tex]
During metamorphism, what is the major effect of chemically active fluids?
a. increase the pressures in deeply buried, regional-metamorphic zones
b. aid in the movement of dissolved silicate ions and facilitate growth of the mineral grains
c. prevent partial melting so solid rocks can undergo very high temperature regional metamorphism
d. facilitate the formation of schistosity and gneissic banding in hornfels and slates
Final answer:
Chemically active fluids during metamorphism help in the transfer and movement of ions, facilitating the growth and reorganization of mineral grains in the rock.
Explanation:
During metamorphism, chemically active fluids primarily aid in the movement of dissolved silicate ions and facilitate growth of the mineral grains. These fluids can enhance the transfer of ions, which allows for the reorganization of atoms and the growth of new mineral structures within the rock. The presence of chemically active fluids during metamorphism can therefore play a critical role in the transformation of a rock's mineral composition and texture.
The sun radiates energy. Does the earth similarly radiate energy? If so, why can’t we see the radiant energy from the earth?
1. Yes, the earth radiates energy, but only in the sense that the energy is reflected sunlight, which we can’t see from the earth’s surface.
2. No, the earth does not similarly radiate energy.
3. Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the absolute temperature of the radiator, the wavelength of the radiation is far too long for us to see.
4. Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the temperature of the radiator in degrees Kelvin, the frequency of the radiation from the earth is in the ultraviolet range (so that.
Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the absolute temperature of the radiator, the wavelength of the radiation is far too long for us to see.
Answer: Option C
Explanation:
Radiations or light rays are the basic name for electromagnetic energy packets travelling through space. It goes extremely quick (multiple times around the earth in one second) and can go through a vacuum. It needn't bother with material to travel in.
It has numerous structures, including visible light, infrared (IR), bright (UV), X-beams, microwaves, and radio waves. These are a no different structure of energy, just with various frequencies and measures of energy. Various frequencies of radiation communicate with an issue in an unexpected way, which causes them to appear to be more changed to us than they truly are.
What would be the wavelength of the fundamental and first two overtones produced by an organ pipe of length L that is closed at one end and open at the other?
A) L, ½ L, ¼ L
B) ½ L, ¼ L, 1/6 L
C) 2L, L, ½ L
D) 4L, 2L, L
E) 4L, 4/3 L, 4/5 L
Answer:
option E
Explanation:
given,
organ pipe of length L
using formula ,
[tex]L = (2n - 1)\dfrac{\lambda}{4}[/tex]
n is the number of nodes
[tex]\lambda= \dfrac{4L}{(2n - 1)}[/tex]
now at n = 1
[tex]\lambda_1= \dfrac{4L}{(2(1) - 1)}[/tex]
[tex]\lambda_1= 4L[/tex]
now at n = 2
[tex]\lambda_2= \dfrac{4L}{(2(2) - 1)}[/tex]
[tex]\lambda_2= \dfrac{4}{3}L[/tex]
now at n = 2
[tex]\lambda_3= \dfrac{4L}{(2(3) - 1)}[/tex]
[tex]\lambda_3= \dfrac{4}{5}L[/tex]
The correct answer is option E
Answer:
The correct answer is option E
Explanation:
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a 3.00-m^2 window that is 0.635 cm thick (1/4 in) if the temperatures of the inner and outer surfaces are 5.00ºC and −10.0ºC, respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation. (answer in ×10^{3} W)
Answer:
[tex]5.9527559\times 10^3\ W[/tex]
Explanation:
Q = Heat
t = Thickness = d = 0.635 cm
[tex]k_g[/tex] = Heat conduction coefficient of glass = 0.84 W/m °C (general value)
[tex]\Delta T[/tex] = Change in temperature
A = Area = 3 m²
Power is given by
[tex]P=\frac{dQ}{dt}=\frac{kA\Delta T}{d}\\\Rightarrow P=\frac{k_gA\Delta T}{d}\\\Rightarrow P=\frac{0.84\times 3(5-(-10))}{0.00635}\\\Rightarrow P=5.9527559\times 10^3\ W[/tex]
The rate of conduction in watts through the window is [tex]5.9527559\times 10^3\ W[/tex]
The rate of heat conduction out of a window is calculated by using Fourier's law. Substituting the given values into Fourier's law yields a heat transfer rate of around 1.97 × 10⁴ W. This represents a significant amount of heat loss through the windows.
Explanation:The heat conduction rate can be calculated using Fourier's law, which states that the rate of heat transfer per time, or heat flux, is proportional to the temperature gradient. The formula for this law is Q = k*A*(T1-T2)/d, where Q is the heat transfer rate in watts, k is the thermal conductivity, A is the surface area in square meters, T1 and T2 are the temperatures of the two surfaces, and d is the thickness of the material. Given that glass has a thermal conductivity of about 0.84 W/mºC, when we substitute these values into Fourier's law, we obtain:
Q = 0.84 W/mºC * 3.00 m² * (5.00ºC - (-10.0ºC))/0.635 cm, whereby this yields around 1.97 × 10⁴ W.
This indicates that a significant amount of heat is being lost through the windows, contributing to the chilling of the air in the room.
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A uranium nucleus 238U may stay in one piece for billions of years, but sooner or later it decays into an α particle of mass 6.64×10−27 kg and 234Th nucleus of mass 3.88 × 10−25 kg, and the decay process itself is extremely fast (it takes about 10−20 s). Suppose the uranium nucleus was at rest just before the decay. If the α particle is emitted at a speed of 8.57×106 m/s, what would be the recoil speed of the thorium nucleus? Answer in units of m/s.
Answer:
1.47*10^5m/s
Explanation:
Using law of conservation of momentum
Since the uranium was initially at rest then the total momentum (did not change) is conserved and its initial value is zero ; 0= M1*V1+M2V2
-M1V1 = M2V2 where V1 is the velocity of the alpha particle and m1 is its mass and M2 is the mass of the thorium and V2 is its velocity.
- 6.64*10^-27*8.56*10^6 = 3.88*V2
V2 = -5.69*10^-20/3.88*10^-25 = -1.47*10^5m/s the negative sign mean the vector velocity of the nucleus point in opposite direction to the particle.
The recoil speed of the thorium nucleus after uranium-238 decays is approximately 0.14717 x 10^6 m/s. This is calculated using the law of conservation of momentum, where the initial momentum of the uranium-238 nucleus at rest is equal to the combined momentum of the emitted α-particle and the resulting thorium nucleus.
Explanation:The recoil speed of the thorium nucleus can be calculated using the law of conservation of momentum. Before the decay, the momentum of the uranium-238 nucleus (the parent nuclide) is zero, as it's at rest. After the decay, the momentum has to remain zero for the system, including the α-particle and the daughter nucleus thorium-234.
It's given that the mass of the α-particle is 6.64×10−27 kg and it's emitted at a speed of 8.57×10^6 m/s. So, the momentum of the α-particle is the product of its mass and velocity, which is then equated to the momentum of the Thorium-234, which is its mass times its velocity (recoil speed).
Setting the two momentums equal to each other and solving for the recoil speed of thorium (V) gives us: (6.64×10−27 kg * 8.57×10^6 m/s) / 3.88 × 10−25 kg = V. Calculating this gives a recoil speed of about 0.14717 * 10^6 m/s for the thorium nucleus.
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s) A body of mass 2 kilograms moves on a circle of radius 3 meters, making one revolution every 5 seconds. Find the magnitude of the centripetal force acting on the body. (Be sure to correctly state the units of the force
Answer:
Centripetal force acting on the body = 9.47 N
Explanation:
Mass of body, m = 2 kg
Radius, r = 3 m
It makes one revolution in 5 seconds.
Period, T = 5 s
[tex]\texttt{Angular velocity, }\omega =\frac{2\pi }{T}=\frac{2\pi }{5}=1.256rad/s[/tex]
Centripetal force, F = mrω²
F = 2 x 3 x 1.256² = 9.47 N
Centripetal force acting on the body = 9.47 N
The mechanical vibrations triggered by sound waves are transduced into neural impulses by
Sound waves cause the eardrum to vibrate, which then send these vibrations to the middle ear and then to the cochlea of the inner ear. Herein, the vibrations are converted by hair cells on the basilar membrane into neural impulses which are transmitted to the brain and perceived as sound.
Explanation:The mechanical vibrations, also referred to as sound waves, reach the outer ear and are transferred to the ear canal where they cause the tympanic membrane, or eardrum, to vibrate. These vibrations are passed on to the three bones of the middle ear, namely the malleus, incus, and stapes. The stapes then transmits these vibrations to a structure known as the oval window, which is the outermost structure of the inner ear. Herein lies the cochlea, a spiral-shaped structure filled with fluid, and the vibrations from the oval window create pressure waves within this fluid. Within the cochlea, a structure known as the basilar membrane, which hosts receptor hair cells, is mechanically stimulated by these pressure waves. When the hair cells 'bend', this initiates a process of mechanical transduction, wherein the mechanical vibrations from sound waves are transformed, or transduced, into electrical signals known as neural impulses. These impulses are then transmitted to the brain via the cochlear nerve, whereby they are perceived as sound.
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The mechanical vibrations of sound waves are transduced into neural impulses by the hair cells in the inner ear.
The mechanical vibrations of sound waves are transduced into neural impulses by specialized cells in the inner ear called hair cells. These hair cells are located within the cochlea, a spiral-shaped organ responsible for auditory processing. When sound waves reach the inner ear, they cause the hair cells to bend. This bending motion activates ion channels on the hair cell membrane, leading to the generation of electrical signals or neural impulses.
These signals are then transmitted through the auditory nerve to the brain, where they are interpreted as sound. The hair cells' ability to transduce mechanical vibrations into neural impulses is a crucial step in the auditory process, allowing us to perceive and understand various sounds in our environment, from music to speech and other auditory stimuli. This complex process enables our sense of hearing and is essential for our communication and interaction with the world around us.
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Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.
Answer:
t = 83.93 nm
Explanation:
given,
blue light wavelength (λ)= 470 nm
refractive index of oil (μ)= 1.40
minimum thickness of an oil slick on water = ?
using constructive interference formula
now,
[tex]2 \mu t = (m + \dfrac{1}{2})\lambda[/tex]
where, t is the thickness of the oil slick
m = 0,1,2
for minimum thickness m = 0
[tex]2\times 1.40\times t = (0 + 0.5)\times 470[/tex]
[tex]2.8\times t = 235[/tex]
[tex]t = \dfrac{235}{2.8}[/tex]
t = 83.93 nm
minimum thickness of an oil slick on water = t = 83.93 nm
The minimum thickness of an oil slick should be 83.93 nm
Important information:Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40
Calculation of the minimum thickness:[tex]2\times 1.40 \times t= (0 + 0.5) \times 470\\\\28 \times t = 235[/tex]
t = 83.93 nm
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Methane gas absorbs red light, and methane clouds reflect blue light, giving Uranus and Neptune their distinctive blue colors. Why do Uranus and Neptune have methane clouds, but Jupiter and Saturn do not
Answer:
Jupiter and Saturn have high temperatures and they have very large gravitational pull compared to Uranus and Neptune.
Explanation:
Jupiter and Saturn have high densities meaning any methane on them is dragged down to their hot surface which prevents methane from ever forming clouds.
Answer:
Jupiter and Saturn have high temperatures and they have very large gravitational pull compared to Uranus and Neptune.
Explanation:
A jet plane at take-off can produce a sound of intensity 9.70 W/m2 at 30.3 m away. But you prefer the tranquil sound of normal conversation, which is 1.0μW/m2. Assume that the plane behaves like a point source of the sound.
a. What is the closest distance you should live from the airport runway to preserve your peace of mind?
b. What intensity from the jet does your friend experience if she lives twice as far from the runway as you do?
c. What power of sound does the jet produce at take-off?
Assume that the plane behaves like a point source of the sound.The answers are a. Closest distance is approximately 29820 meters. b. Your friend experiences an intensity of 0.25 μW/m². c. The jet produces approximately 112 kW of sound power at take-off.
Given the sound intensity[tex]\(I = 9.70 \text{ W/m}^2\)[/tex]at a distance [tex]\(d = 30.3 \text{ m}\)[/tex], and the desired intensity[tex]\(I_{desired} = 1.0 \mu \text{W/m}^2 = 1.0 \times 10^{-6} \text{ W/m}^2\),[/tex]we can use the inverse square law for sound intensity:
[tex]\[ I \propto \frac{1}{d^2} \][/tex]
(a) To find the closest distance [tex]\(d_{desired}\)[/tex]:
[tex]\[ \frac{I}{I_{desired}} = \left( \frac{d_{desired}}{d} \right)^2 \]\[ \frac{9.70}{1.0 \times 10^{-6}} = \left( \frac{d_{desired}}{30.3} \right)^2 \]\[ d_{desired} \approx 30.3 \times \sqrt{9.70 \times 10^6} \approx 29820 \text{ m} \][/tex]
(b) The intensity at twice the distance[tex]\(2d_{desired}\)[/tex]:
[tex]\[ I_{friend} = \frac{I_{desired}}{4} = 0.25 \times 1.0 \mu \text{W/m}^2 = 0.25 \mu \text{W/m}^2 \][/tex]
(c) The power of the jet can be found using:
[tex]\[ I = \frac{P}{4\pi d^2} \]\[ P = I \times 4\pi d^2 \]\[ P = 9.70 \times 4\pi \times (30.3)^2 \approx 1.12 \times 10^5 \text{ W} \][/tex]
Technician A says that all FWD vehicles use adjustable front wheel bearings. Technician B says that most FWD vehicles use sealed non-adjustable front wheel bearings. Which technician is correct?
Answer:
Explanation:
Technician B only.
Suppose a star with radius 8.50 x 10⁸ m has a peak wavelength of 685 nm in the spectrum of its emitted radiation. (a) Find the energy of a photon with this wavelength. (b) What is surface temperature of the star? (c) At what rate is energy emitted from the star in the form of radiation? Assume the star is a blackbody (e = 1). (d) Using the answer to part (a), estimate the rate at which photons leave the surface of the star.
Answer:
a) Energy of Photon = 2.9 * 10^ -19 J
A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density of (a) the magnetic field and (b) the electric field at the surface of the wire. The wire diameter is 2.8 mm, and its resistance per unit length is 3.7 Ω/km.
a. The energy density of the magnetic field ([tex]\(u_B\)[/tex]) at the surface of the wire is [tex]\(3.08875 \times 10^{-7} \, \text{J/m}^3\)[/tex]
b. The energy density of the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire is [tex]\(3.74883 \times 10^{-3} \, \text{J/m}^3\).[/tex]
To calculate the energy density of the magnetic field ([tex]\(u_B\)[/tex]) and the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire, we first need to find the magnetic field strength (B) and the electric field strength (E) at that point.
Given:
- Current (I) = 11 A
- Wire diameter (d) = 2.8 mm = 2.8 x 10^-3 m
- Resistance per unit length (R) = 3.7 Ω/km = 3.7 x 10^3 Ω/m
We can calculate the radius (r) of the wire:
[tex]\[ r = \frac{d}{2} = \frac{2.8 \times 10^{-3}}{2} = 1.4 \times 10^{-3} \, \text{m} \][/tex]
(a) Magnetic Field Energy Density ([tex]\(u_B\)[/tex]):
The magnetic field strength at the surface of the wire (B) is given by Ampère's Law:
[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]
[tex]\[ B = \frac{4\pi \times 10^{-7} \times 11}{2\pi \times 1.4 \times 10^{-3}} \][/tex]
[tex]\[ B = \frac{4 \times 11 \times 10^{-7}}{2 \times 1.4 \times 10^{-3}} \][/tex]
[tex]\[ B = \frac{4 \times 11}{2 \times 1.4} \times 10^{-4} \][/tex]
[tex]\[ B = \frac{44}{2.8} \times 10^{-4} \][/tex]
[tex]\[ B = 15.714 \times 10^{-4} \][/tex]
[tex]\[ B = 1.571 \times 10^{-3} \, \text{T} \][/tex]
The energy density of the magnetic field ([tex]\(u_B\)[/tex]) is given by:
[tex]\[ u_B = \frac{B^2}{2\mu_0} \][/tex]
[tex]\[ u_B = \frac{(1.571 \times 10^{-3})^2}{2 \times 4\pi \times 10^{-7}} \][/tex]
[tex]\[ u_B = \frac{2.471 \times 10^{-6}}{8\pi \times 10^{-7}} \][/tex]
[tex]\[ u_B = \frac{2.471}{8} \times 10^{-6} \][/tex]
[tex]\[ u_B = 0.308875 \times 10^{-6} \][/tex]
[tex]\[ u_B = 3.08875 \times 10^{-7} \, \text{J/m}^3 \][/tex]
(b) Electric Field Energy Density ([tex]\(u_E\)[/tex]):
The electric field strength (E) at the surface of the wire is given by Ohm's Law:
[tex]\[ E = \frac{V}{r} \][/tex]
[tex]\[ E = \frac{IR}{r} \][/tex]
[tex]\[ E = \frac{11 \times 3.7 \times 10^3}{1.4 \times 10^{-3}} \][/tex]
[tex]\[ E = \frac{40.7 \times 10^3}{1.4 \times 10^{-3}} \][/tex]
[tex]\[ E = \frac{40.7}{1.4} \times 10^3 \][/tex]
[tex]\[ E = 29.071 \times 10^3 \][/tex]
[tex]\[ E = 29.071 \times 10^3 \, \text{V/m} \][/tex]
The energy density of the electric field ([tex]\(u_E\)[/tex]) is given by:
[tex]\[ u_E = \frac{\varepsilon_0 E^2}{2} \][/tex]
[tex]\[ u_E = \frac{8.85 \times 10^{-12} \times (29.071 \times 10^3)^2}{2} \][/tex]
[tex]\[ u_E = \frac{8.85 \times 10^{-12} \times 845.96 \times 10^6}{2} \][/tex]
[tex]\[ u_E = \frac{7.49766 \times 10^{-3}}{2} \][/tex]
[tex]\[ u_E = 3.74883 \times 10^{-3} \][/tex]
So, the energy density of the magnetic field ([tex]\(u_B\)[/tex]) at the surface of the wire is [tex]\(3.08875 \times 10^{-7} \, \text{J/m}^3\)[/tex] and the energy density of the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire is [tex]\(3.74883 \times 10^{-3} \, \text{J/m}^3\).[/tex]
Block 1, of mass m1 = 0.500 kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0 ∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.400, an acceleration of magnitude a = 0.500 m/s2 is observed for block 2.
Find the mass of block 2, m2.
Express your answer numerically in kilograms.
Answer:
m2 = 0·52 kg
Explanation:
As the pulley is massless and frictionless the tension in the string on both sides of the pulley will be the same
Given that both have same acceleration a = 0·5 m/s²
Forces acting on mass m1 is tension and force of gravity
You can observe the direction of forces acting on two blocks in the file attached
And in the file attached the force of gravity acting on mass m2 is resolved into two perpendicular components of which one acts along the wedge and the other perpendicular to the wedge
Let the tension in the string be T N
And the frictional force acting on mass m2 is μ×N as sliding is taking place
By applying Newton's second law to the block of mass m1
m1×g - T = m1 × a
⇒ T = m1×(g - a)
⇒T = 0·5×(9·8-0·5)
⇒ T= 4·65 N
Let the normal reaction acting on mass m2 be N
By applying Newton's second law to the block of mass m2 in the direction perpendicular to the wedge
we get
N = m2×g×cosθ
By applying Newton's second law to the block of mass m2 along the wedge
T - μ×N - m2×g×sinθ = m2×a
Substitute N = m2×g×cosθ in the above equation
T - μ×m2×g×cosθ - m2×g×sinθ = m2×a
⇒ T = m2 × ( μ×g×cosθ + g×sinθ + a)
By the substituting the corresponding values
4·65 = m2 × 8·8
⇒ m2 = 0·52 kg
The mass of block 2, m2.is mathematically given as
m2 = 0·52 kg
What is the mass of block 2, m2.?Question Parameter(s):
mass m1 = 0.500 kg , an angle = 30.0 ∘f μ = 0.400,a = 0.500 m/s2Generally, the equation for the Newton's second law is mathematically given as
m1×g - T = m1 × a
Therefore
T = m1*(g - a)
T = 0·5*(9·8-0·5)
T= 4·65 N
Therefore for
T - u*N - m2*g×sin[tex]\theta[/tex] = m2*a
4·65 = m2 × 8·8
m2 = 0·52 kg
In conclusion, the mass of block 2
m2 = 0·52 kg
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Most cells cannot harness heat to perform work because
Answer:
temperature is usually uniform throughout a cell or temperature is usually uniform to do work.
Explanation:
Heat (thermal energy) is a kinetic energy. It s connected with the random movement of the atoms or molecules. The temperature is usually uniform throughout a cell, so most of the cells cannot harness heat to perform work.
Cells are unable to harness heat to perform work due to the second law of thermodynamics, which results in energy being lost in a form that is unusable, often as heat, during transfers and transformations. This heat energy is essentially lost to the cell for performing work. And as entropy increases, less energy becomes available for work.
Explanation:Most cells cannot harness heat to perform work because of the second law of thermodynamics. This law states that all energy transfers and transformations are never completely efficient, with some amount of energy being lost in a form that is unusable, commonly as heat energy. Strictly speaking, heat energy is defined as the energy transferred from one system to another that is not doing work.
For example, during cellular metabolic reactions, a portion of the energy is lost as heat energy. Despite it contributing to maintaining the body temperature of warm-blooded creatures, it is essentially lost to the cell for performing work. This demonstrates how the second law of thermodynamics makes the tasks of cells obtaining, transforming, and using energy to do work more difficult than they appear.
Taking into account the entropy, less and less energy in the universe is available to do work as entropy increases. Eventually, as all fuels are exhausted and temperatures equalize, it will become impossible for heat engines to function, or for work to be done. Hence, cells can't harness heat to perform work due to the inefficiencies described by the second law of thermodynamics.
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Two 2.0 kg bodies, A and B, collide. The velocities before the collision are
v→A=(15i^+30j^) m/s
and
v→B=(−10i^+5.0j^) m/s
. After the collision,
v→A=(−5.0i^+20j^) m/s
. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
Answer:
Explanation:
Given
mass of body A [tex]m_a=2 kg[/tex]
mass of body [tex]m_b=2 kg[/tex]
Velocity before Collision is [tex]u_a=15\hat{i}+30\hat{j}[/tex]
[tex]u_b=-10\hat{i}+5\hat{j} m/s[/tex]
after collision [tex]v_a=-5\hat{i}+20\hat{j} m/s[/tex]
let [tex]v_b_x[/tex] and [tex]v_b_y[/tex] velocity of B after collision in x and y direction
conserving momentum in x direction
[tex]m_a\times 15+m_b\times (-10)=m_a\times (-5)+m_b\times (v_b_x)[/tex]
as [tex]m_a=m_b[/tex] thus
[tex]15-10=-5+v_b_x[/tex]
[tex]v_b_x=10 m/s[/tex]
Conserving momentum in Y direction
[tex]m_a\times 30+m_b\times 5=m_a\times 20+m_b\times (v_b_y)[/tex]
[tex]30+5=20+v_b_y[/tex]
[tex]v_b_y=15 m/s[/tex]
thus velocity of B after collision is
[tex]v_b=10\hat{i}+15\hat{j}[/tex]
(b)Change in total Kinetic Energy
Initial Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{15^2+30^2})^2=1125 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+5^2})^2=125 J[/tex]
Total initial Kinetic Energy =1250 J
Final Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{5^2+20^2})^2=425 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+15^2})^2=325 J[/tex]
Final Kinetic Energy [tex]=425+325=750 J[/tex]
Change [tex]\Delta K.E.=1250-750=500 J[/tex]
The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered what very significant fact about the spectra produced by hot gases, such as elements heated in a flame?
The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.
Explanation:A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.
Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.
The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.
What is the initial drag force for a golf ball with the following conditions? Frontal area: 14 cm2 (remember to convert to m2), air density: 1.204 kg/m3, initial velocity: 50 m/s, coefficient of drag: 0.3
Answer:
F= 0.6321 N
Explanation:
Given
Area ,A= 14 cm²
Density ,ρ = 1.204 kg/m³
Velocity ,v= 50 m/s
Drag coefficient ,C=0.3
The drag force on the golf given as
[tex]F=\dfrac{1}{2}\rho CAv^2[/tex]
Now by putting the values
[tex]F=\dfrac{1}{2}\rho CAv^2[/tex]
[tex]F=\dfrac{1}{2}\times 0.3\times 1.204 \times (14\times 10^{-4})\times 50^2\ N[/tex]
F= 0.6321 N
Therefore force due to drag on the golf is 0.6321 N
A copper-constantan thermocouple generates a voltage of 4.75 x 10-3 volts when the temperature of the hot junction is 110 °C and the reference junction is kept at 0 °C. If the voltage is proportional to the difference in temperature between the junctions, what is the temperature in degrees Celsius of the hot junction when the voltage is 1.76 x 10-3 volts?
Answer:
The temperature of the hot junction is 40.7°C.
Explanation:
Given that,
Voltage [tex]V=4.75\times10^{-3}\ volts[/tex]
Voltage [tex]V'=1.76\times10^{-3}\ volt[/tex]
Temperature of hot junction = 110°C
If the voltage is proportional to the difference in temperature between the junctions,
We need to calculate the temperature of the hot junction
Using formula of temperature
[tex]\dfrac{V}{V'}=\dfrac{\Delta T}{\Delta T}[/tex]
[tex]\dfrac{V}{V'}=\dfrac{T_{2}-T_{1}}{T_{2}-T_{1}}[/tex]
Here,T₁=0°C
[tex]\dfrac{V}{V'}=\dfrac{110}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{4.75\times10^{-3}}{1.76\times10^{-3}}=\dfrac{110}{T_{2}}[/tex]
[tex]T_{2}=\dfrac{110\times1.76\times10^{-3}}{4.75\times10^{-3}}[/tex]
[tex]T_{2}=40.7^{\circ}C[/tex]
Hence, The temperature of the hot junction is 40.7°C.
Final answer:
To determine the temperature of the hot junction in a copper-constantan thermocouple when the voltage is 1.76 x 10-3 volts, we can set up a proportion between the voltages and the temperature difference.
Explanation:
To determine the temperature of the hot junction when the voltage is 1.76 x 10-3 volts, we can use the proportionality between voltage and temperature difference. We know that the voltage generated by the thermocouple is directly proportional to the difference in temperature between the junctions. So, we can set up a proportion:
(4.75 x 10-3 volts) / (110 °C - 0 °C) = (1.76 x 10-3 volts) / (x °C - 0 °C)
By cross-multiplying and solving for x, we can find the temperature of the hot junction. The result is x = 41.5 °C.
How long will it take a 2.3"x10^3 kg truck to go from 22.2 m/s to a complete stop if acted on by a force of -1.26x10^4 N.What would be it's stopping distance?
The stopping distance is 45.0 m
Explanation:
First of all, we find the acceleration of the truck, by using Newton's second law:
[tex]F=ma[/tex]
where
[tex]F=-1.26\cdot 10^4 N[/tex] is the net force on the truck
[tex]m=2.3\cdot 10^3 kg[/tex] is the mass of the truck
a is its acceleration
Solving for a,
[tex]a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2[/tex]
where the negative sign means the acceleration is opposite to the direction of motion.
Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity of the truck
u = 22.2 m/s is the initial velocity
[tex]a=-5.48 m/s^2[/tex] is the acceleration
s is the stopping distance
And solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m[/tex]
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The science that uses principles of physics to understand the human machine is called
Answer:
Biomechanics
Explanation:
Biomechanics is a branch of biophysics that uses principles of physics to study and understand movement in organisms using the laws of mechanics. One of the applications of Biomechanics is its usage in sport and exercise activities. Biomechanics is used to study body movement of athletes during sport or exercise activities to improve performance of athletes and make the activities better and safer.
What kinds of forces hold ionic solids together? Check all that apply. Check all that apply. dispersion forces metallic bonds covalent bonds electrostatic attraction dipole-dipole forces hydrogen bonds
Answer:
electrostatic attraction.
Explanation:
As we know those atom donate the electron gets positively charged and those atoms gains the electron gets negatively charged.The charge on the atoms are the responsible for the force in the ionic bond.
We know that force between two charge particle is known as electrostatics force.If charge particle is having same sign charge then it will be repulsive force and If charge particle is having opposite sign charge then it will be attraction force.
Therefore force in the ionic solid is electrostatic attraction.
Final answer:
Ionic solids are primarily held together by electrostatic attraction, with dispersion forces playing a possible but minor role. Metallic bonds, covalent bonds, dipole-dipole forces, and hydrogen bonds do not typically contribute to the structure of ionic solids.
Explanation:
The kinds of forces that hold ionic solids together include electrostatic attraction and possibly dispersion forces if there are also nonpolar parts or induced dipoles present. Electrostatic attraction, also known as ionic bonding, is the main force that holds together the ions in an ionic solid, resulting from the attraction between positively charged cations and negatively charged anions. However, dispersion forces can also play a lesser role when nonpolar regions or molecules are adjacent to the ionic components, though they are not the primary force in ionic solids.
Metallic bonds, covalent bonds, dipole-dipole forces, and hydrogen bonds are not typically involved in the structure of ionic solids. Metallic bonds are specific to metal lattices, covalent bonds connect atoms within molecules or network solids, dipole-dipole forces and hydrogen bonds are types of intermolecular forces mainly significant in molecular compounds, not ionic compounds.
An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ above the horizontal. The block is displaced by 19.9 m, and the coefficient of kinetic friction is 0.209.
a. Find the work done by the 156 N force. The acceleration of gravity is 9.8 m/s^2
b. Find the magnitude of the work done by the orce of friction
c. What is the sign of the work done by the frictional force?
d. Find the work done by the normal force
Answer:
a) W = 2635.56 J
b) Wf = 423.27 J
c) c) The Sign of the work done by the frictional force (Wf) is negative (-)
d) W=0
Explanation:
Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .
The formula for calculate the work is :
W = F*d*cosα
Where:
W : work in Joules (J)
F : force in Newtons (N)
d: displacement in meters (m)
α :angle that form the force (F) and displacement (d)
Known data
m = 18.8 kg : mass of the block
F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal
μk= 0.209 : coefficient of kinetic friction between the cart and the surface
g = 9.8 m/s²: acceleration due to gravity
d = 19.9 m : displacement of the block
Forces acting on the block
We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.
W: Weight of the cart : In vertical direction downaward
N : Normal force : In vertical direction the upaward
F : Force applied to the block
f : Friction force: In horizontal direction
Calculated of the weight of the block
W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N
x-y components of the force F
Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N
Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n
Calculated of the Normal force
Newton's second law for the block in y direction :
∑Fy = m*ay ay = 0
N-W+Fy= 0
N-184.24+82,44= 0
N = 184.24-82,44
N = 101.8 N
Calculated of the kinetic friction force (fk):
fk = μk*N = (0.209)*( 101.8)
fk = 21.27 N
a) Work done by the F=156N.
W = (Fx) *d *cosα
W = (132.44 )*(19.9)(cos0°) (N*m)
W = 2635.56 J
b) Work done by the force of friction
Wf = (fk) *d *cos(180°)
Wf = (21.27 )*(19.9) (-1) (N*m)
Wf = - 423.27 J
Wf = 423.27 J :magnitude
c) The Sign of the work done by the frictional force is negative (-)
d) Work done by the Normal force
W = (N) *d *cos(90°)
W = (101.8 )*(19.9) (0) (N*m)
W = 0
The work done by the 156 N force is 2630.77 J. The magnitude of the work done by the friction force is 743.14 J. The work done by the normal force is zero.
Explanation:a. The work done by a force is calculated by multiplying the force applied by the displacement in the direction of the force. In this case, the force is applied at an angle of 31.9 degrees above the horizontal, so we need to find the component of the force in the horizontal direction.
Fx = F * cos(θ) = 156 N * cos(31.9°) = 132.3 N
The work done is given by W = Fx * d = 132.3 N * 19.9 m = 2630.77 J
b. The magnitude of the work done by the force of friction can be calculated using the formula:
Wfriction = friction force * displacement = μ * m * g * d, where μ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and d is the displacement. Substituting the given values:
Wfriction = 0.209 * 18.8 kg * 9.8 m/s2 * 19.9 m = 743.14 J
c. The work done by the frictional force is negative, indicating that it acts against the motion of the block.
d. The work done by the normal force is zero because the displacement of the block is perpendicular to the direction of the normal force.
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A massless spring with spring constant 16.4 N/m hangs vertically. A body of mass 0.193 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find (a) how far below the initial position the body descends, and the (b) frequency and (c) amplitude of the resulting SHM.
Answer:
(A) 0.2306 m
(B) 1.467 Hz
(C) 0.1152 m
Explanation:
spring constant (K) = 16.4 N/m
mass (m) = 0.193 kg
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) force = Kx, where x = extension
mg = Kx
0.193 x 9.8 = 16.4x
x = 0.1153 m
now the mass actually falls two times this value before it gets to its equilibrium position ( turning point ) and oscillates about this point
therefore
2x = 0.2306 m
(B) frequency (f) = \frac{1}{2π} x [tex]\sqrt{\frac{k}{m}}[/tex]
frequency (f) = \frac{1}{2π} x [tex]\sqrt{\frac{16.4}{0.193}}[/tex]
frequency = 1.467 Hz
(C) the amplitude is the maximum position of the mass from the equilibrium position, which is half the distance the mass falls below the initial length of the spring
= \frac{0.2306}{2} = 0.1152 m
Final answer:
The mass descends by 0.1154 meters, the frequency of SHM is about 1.83 Hz, and the amplitude of the oscillation is 0.1154 meters.
Explanation:
When a mass of 0.193 kg is hung on a vertical massless spring of spring constant 16.4 N/m, the spring is stretched due to the weight of the mass. The distance that the body descends is equal to the distance at which the spring force balances the gravitational force on the mass. This can be calculated by using Hooke's Law (F = kx), where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. The force due to gravity on the mass is given by W = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s2).
(a) To find the descent (x), set the spring force equal to the gravitational force: mg = kx, which yields x = mg/k. Plugging in the values yields x = (0.193 kg)(9.8 m/s2) / 16.4 N/m, which gives x = 0.1154 meters.
(b) The frequency of the resulting SHM (Simple Harmonic Motion) can be determined by the formula f = (1/2π) √(k/m). Substituting the values, f = (1/2π) √(16.4 N/m / 0.193 kg), which gives f ≈ 1.83 Hz.
(c) The amplitude of SHM is equal to the maximum displacement from the equilibrium position, which is the same as the distance calculated in part (a), hence the amplitude is 0.1154 meters.