Answer:
Solution
Given:
initial velocity (u)=0m/s (because an object
starts from the rest)
time (t)=3sec
diatance travelled(s)=14m(16-2=14)
acceleration (a)=?
Now,
According to the formula
s=ut+1/2at^2
or,14 =0×3+1/2×a×3^2
or, 14=0+a/2×9
or, 14=9a/2
or, 9a=28
or, a=28/9
or, a=3.1 m/s^2
Therefore, the acceleration of the object is 3.1m/s^2 ans.
Will mark as BRAINLIEST.......
The position vector is given by vector r= 5t² I cap + 2 t³ j cap + 2 k cap. Find it's velocity and acceleration at t=2s.
Answer:
We have the position vector given in terms of time t. r(t) = t^3*i + t^2*j
To find the velocity vector we have to differentiate r(t) with respect to time.
r'(t) = 3t^2*i + 2t*j
The vector representing acceleration is the derivative of the position vector
r''(t) = 6t*i + 2*j
When time t = 2.
The velocity vector is 3*2^2*i + 2*2*j
=> 12*i + 4*j
The speed is the absolute value of the velocity vector or sqrt(12^2 + 4^2) = sqrt (144 + 16) = sqrt 160
The acceleration vector is 6*2*i + 2*j
=> 12*i + 2*j
The required acceleration at t=2 is 12*i + 2*j and the speed is sqrt 160.
Explanation:
Can I have thx and brainliest?
(b) A cylinder of cross-sectional area 0.65m2 and
height 0.32m has a mass of 2. Ikg. If there is a
cavity inside, find the volume of the cavity.
(Density of cylinder = 11.053 kg/m^3)
Answer:
The volume of the cavity is 0.013m^3
Explanation:
To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:
Step one:
Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.
Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]
Step two:
From the volume of the cylinder, we can get the radius of the cylinder.
[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]
Step three:
From the cross-sectional area, we can obtain the radius of the cavity.
Let the radius of the cavity be = r, while the radius of the cylinder be = R
CSA of cavity =
[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]
Step Four:
calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]
Recall that the cavity has the same height as the original cylinder
[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]
Suppose an experiment is designed to test the durability of batteries in different conditions. All of the batteries tested are double-A (AA) Brand X. All sets of batteries are preconditioned in different environmental conditions for exactly 168 hours (1 week).
Set 1: 0°C (freezing point of water)
Set 2: 24°C (approximately room temperature)
Set 3: 37°C (approximately body temperature)
The batteries are then continuously used to power identical mechanical drummer toys. As long as the toy keeps drumming the battery is considered functional. The drumming time for each toy is measured as an indication of battery durability. In this experiment, which condition is not controlled?
A.) temperature
B.) brand of batteries
C.) test for durability
D.) type of battery (battery size)
Answer:
I assume its c. Since its talking about testing.
Explanation:
Answer:
The answer is test of durability
Explanation:
Can someone please illustrate how the refracted ray will look like?
Answer
As the angle of incidence increases in Figure 2.8, a point is finally reached where the refracted ray does not emerge at the second layer but lie along the interface. This particular angle of incidence at which the angle of refraction is 90° and the refracted ray lies along the interface is known as the critical angle. At and beyond the critical angle, there is no transmitted ray and therefore a very high reflected ray will be recorded .
Therefore,
sinθisin90=Vp1Vp2
But, sin 90 = 1.
At critical angle,
sinθcritical=Vp1Vp2
A critical refracted wave travels along the interface between layers and is refracted back into the upper layer at the critical angle. The waves refracted back into the upper layer are called head waves or first-break refractions because at certain distances from a source, they are the first arriving energy. Recorded first-break refraction is shown in Figure 2.10.
Note that these first-break refractions can give us important information about the shallow velocities on land seismic data.
Note also that seismic data are acquired in such a way that reflections from horizons of interest are in the pre-critical region, even at the farthest offset in the data.
In reality, part of the seismic energy arriving at an interface is transmitted and refracted, and another part of the energy is reflected at that same interface. Given that there are many reflectors in the subsurface, there will be many paths from source to receiver, each of them with a different travel time. The proportion of energy reflected depends on the material properties of the two bounding layers and on the angle of incidence
Describe the motion of water waves.
Answer:
Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.
an object weights 0.250 kgf in air 0.150 in water and 0.125 in an oil.find out the density of the object and the oil
Answer:
1) The density of the object = 2500 kg/m³
2) The density of the oil = 1250 kg/m³
Explanation:
1) The information relating to the question are;
The mass of the object in air = 0.250 kgf
The mass of the object in water = 0.150 kgf
The mass of the object in the oil = 0.125 kgf
By Archimedes's principle, we have;
The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced
The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf
∴ The weight of the water displaced = 0.1 kgf
Given that the object is completely immersed in the water, we have;
The volume of the water displaced = The volume of the object
The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)
The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³
The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³
The density of the object = 2500 kg/m³
2) Whereby the mass of the object in the oil = 0.125 kgf
The upthrust of the oil = The weight of the oil displaced
The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil
The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf
The weight of the oil displaced = The upthrust of the oil
Given that the volume of the oil displaced = The volume of the oil, we have;
The volume of the oil displaced = 0.0001 m³
The mass of the 0.0001 m³ = 0.125 kg
Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.
The density of the oil = 1250 kg/m³.
A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 1.00 cm thick flat piece of crown glass and back to air again. The beam strikes at an incident angle of 30 degrees. (a) At what angles do the two colors emerge
Answer:
The color blue emerges at 19.16° and the color red emerges at 19.32°.
Explanation:
The angle at which the two colors emerge can be calculated using the Snell's Law:
[tex]n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})[/tex]
Where:
n₁ is the refractive index of the incident medium (air) = 1.0003
n₂ is the refractive index of the refractive medium:
blue light in crown glass = 1.524
red light in crown glass = 1.512
θ₁ is the angle of the incident light = 30°
θ₂ is the angle of the refracted light
For the red wavelengths we have:
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.512}) = 19.32 ^{\circ} [/tex]
For the blue wavelengths we have:
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.524}) = 19.16 ^{\circ} [/tex]
Therefore, the color blue emerges at 19.16° and the color red emerges at 19.32°.
I hope it helps you!
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
A car is moving on straight highway with a speed of 108 km/h.
Answer:
5.3333 sec
Explanation:
initial speed: u = 108km/hr or 30 m/s
final speed: v = 0m/s
distance travelled: s = 80m
time the car took to stop: = t sec
[tex]v^{2} - u^{2}[/tex] = 2as,
a = ([tex]v^{2} - u^{2}[/tex])/2s
a = (0-900)/160
a = -5.625 [tex]ms^{-2}[/tex]
v = u + at,
t = (v - u)/a
t= (0 - 30)/(-5.625)
t = 5.3333 sec
The marginal cost curve
(a) Lies below the ATC curve when the ATC curve slopes upward.
(b) Intersects the AFC and ATC curves at their respective minimum points.
(c) Lies above the AVC curve when the AVC curve slopes downward.
(d) Intersects the AFC and AVC curves at their respective minimum points.
(e) Intersects the AVC and ATC curves at their respective minimum points
Answer:
c
Explanation:
The marginal cost curve image has been attached from which we can clearly, indicate that
ATC = average total cost
AFC = average fixed cost
AVC = average variable cost.
From the graph we can indicate that the marginal cost curve
(c) Lies above the AVC curve when the AVC curve slopes downward.
Question 5
Calculate the kinetic energy of a car (m - 800 kg) moving at 15 m/s. Write your answer to the nearest whole number in the blank space
provided. Only write the numerical value of the answer without units. Do not leave any space in between numbers.
Answer: Joules
Answer:
90,000Explanation:
[tex]m =800kg\\v = 15\\\\K.E = \frac{1}{2}mv^2\\ K.E= \frac{1}{2} \times 800\times 15^2\\= 400 \times 225\\= 90000 joules\\= 90 kilojoules[/tex]
Self-Check
por Learning
A truck mass 8000 kg and a car a mass 1000
kg are travelling at the same velocity. Which one has greater kinetic energy ? Why?
Answer:
K.E of truck > K.E of car
Explanation:
Mass of the truck = 8000Kg
K.E=[tex]\frac{1}{2} mv[/tex]
K.E =[tex]\frac{1}{2}*8000*v\\ 4000v[/tex]
Mass of the car = 1000 Kg
K.E of the car =[tex]\frac{1}{2}*1000*v\\ 500v[/tex]
Therefore Kinetic energy of the truck is greater than that of the car
observe the virual skateboarder coming down the hill and over the ramp describe how each of newton’s laws of motion can be observed in this action you can choose the dry wet or muddy conditions or some combination of these
Answer:
first part the skater goes down a constant slope ramp, initially he has Newton's second law
pply Newton's third law, the normal is the reaction to the support of the body on the surface
the ramp shoots off. axis becomes zero and therefore with Newton's first law its speed
Explanation:
It is the description of this movement let's write Newton's laws.
* The first law that a body goes at constant speed or zero if the sum of the external forces is zero
* the second law is F = m a
* The third law states that the forces act in pairs of equal magnitude and opposite direction, one applied to each body.
Let's apply these laws to our case
In the first part the skater goes down a constant slope ramp, initially he has Newton's second law when he accelerates from the initial velocity of zero to a terminal velocity.
The expression for this is
Wₓ - fr = ma
W sin θ - μ W cos θ = m a
W = mg
g (sin θ - μ cos) = a
the value of the coefficient of kinetic friction depends on the condition of the surface, dry, wet or muddy
This is Newton's second law
On the Y axis, which is perpendicular to the ramp we have
N- [tex]W_{y}[/tex] = 0
If we apply Newton's third law, the normal is the reaction to the support of the body on the surface, note that it can be different from the weight.
In the second part when he is on the ramp.
In the ramp the skater enters with a speed v, suppose that the ramp has an incline so that the skater can jump, in this case the angle is positive with respect to the axis x
In this case the analysis is similar to the previous one
Newton's second law gives the acceleration of the skater, who when he reaches the end of the ramp shoots off.
At this point the force in the x (horizontal) axis becomes zero and therefore with Newton's first law its speed this axis remains constant and the force in the y axis is the force of gravity and has an acceleration that changes if velocity according to Newton's second law
Answer:look at explanations
Explanation:
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer? The nail exerts an equal force on the hammer in the same direction. The nail exerts a much smaller force on the hammer in the opposite direction. The nail exerts an equal force on the hammer in the opposite direction. The nail exerts a much smaller force on the hammer in the same direction.
Answer:
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.
Hello, I am BrotherEye
Answer:
Answers are
1. "The nail exerts an equal force on the hammer in the opposite direction."
2. "500 N"
3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."
4. "When mass moves closer to the point of rotation, rotational inertia decreases."
5. "The skater spins slower because his rotational inertia has increased."
Explanation:
The pH scale is used to tell if a substance
is an acid or base. Substances with a pH
of 7 are neutral. An acid is anything
below 7 and a base is anything above 7.
Bleach has a pH of about 12. What type of
substance is bleach?
A. base
B. acid
C. neutral
Answer: Bleach is a base
Explanation: If bleach had a pH level of 12, a number above 7, than it is a base. Hope this helps!
A car is driving at 99 km/h, calculate the distance it travels in 70 minutes.
Give your answer in correct SI units rounded to 0 decimal places.
Answer:
The distance the car travels is 115500 m in S.I units
Explanation:
Distance d = vt where v = speed of the car and t = time taken to travel
Now v = 99 km/h. We now convert it to S.I units. So
v = 99 km/h = 99 × 1000 m/(1 × 3600 s)
v = 99000 m/3600 s
v = 27.5 m/s
The speed of the car is 27.5 m/s in S.I units
We now convert the time t = 70 minutes to seconds by multiplying it by 60.
So, t = 70 min = 70 × 60 s = 4200 s
The time taken to travel is 4200 s in S.I units
Now the distance, d = vt
d = 27.5 m/s × 4200 s
d = 115500 m
So, the distance the car travels is 115500 m in S.I units
One glass microscope slide is placed on top of another with their left edges in con- tact and a human hair under the right edge of the upper slide. As a result, a wedge of air exists between the slides. An interference pattern results when monochromatic light is incident on the wedge. What is observed at the left edge of the slides? a. A dark fringe b. A bright fringe c. Impossible to determine
Answer:
A dark fringe
A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net force is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.
Answer:
The answer is The acceleration is double its original value.
Explanation:
It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.Hope this helps....
Have a nice day!!!!
Answer:
The acceleration is half of its original value
Explanation:
element X has two isotopes: X-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and X-29 has an atomic mass of 29.018 and a relative abundance of 17.67%. calculate the atomic mass of element X. show your work
Answer:
27.34 (no unit)
Explanation:
26.975*82.33%+29.018*17.67%
=27.34
Which value would complete the last cell?
(1 point)
3.0
100.0
25.0
4.0
Answer:
4.0
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as
Force (F) = mass (m) x acceleration (a)
F = ma
With the above formula, we can obtain th acceleration of the body as follow:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
F = ma
20 = 5 x a
Divide both side by 5
a = 20/5
a = 4 m/s²
Therefore, the value that will complete the last cell in the question above is 4.
Is there a way for us to control motion
Answer:
They are:
1) change position
2) distract yourself
3) Get fresh air
4) Face the direction you are going.
5) Drink water.
6) Play music.
7) Put your eyes on horizon.
Explanation:
Hope it helps.
State 1 difference between 1 way rotary motion and reversible rotary motion
Answer:
The difference between One-way as well as reversible rotary motion is described below.
Explanation:
Unless rotary motions occur restricted to single direction exclusively (i.e. whether clockwise as well as anti-clockwise only), it is defined as another rotary 'one-way' motion.This motor establishes that continuously variable movement at 360 ° chemically guided is conceivable. The rotating is regulated, and for a particular direction, the biochemical occurrences guiding rotation become incredibly selective.The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive
Answer:
Positive
Explanation:
In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?
Answer:
Rounded to three significant figures:
(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:
A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].MaximaThe path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].
The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].
Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].MinimaThe dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.
First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].
A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s . It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.27 m/s . Suppose the collision is elastic.1. Find the magnitude of the final velocity of the 0.157kg glider.
2. Find the magnitude of the final velocity of the 0.306kg glider.
Answer:
v1 = −2.201946 m/s ( to the left)
v2 = 0.7780534 m/s ( to the right)
Explanation:
Given the following :
Mass of first glider (m1) = 0.149kg
Initial Speed of first glider (u1) = 0.710 m/s
Mass of second glider (m2) = 0.308kg
Initial Speed of second glider (u2) = 2.27m/s
For elastic collision:
m1u1 + mu2u2 = m1v1 + m2v2
Where V1 and v2 = final velocities if the body after collision.
Taking right as positive ; left as negative
u1 = 0.710m/s ; u2 = - 2.27m/s
u1 - u2 = - (v1 - v2)
0.710 - - 2.27 = - v1 + v2
v2 - v1 = 2.98 - - - - (1)
From:
m1u1 + mu2u2 = m1v1 + m2v2
(0.149 * 0.710) + ( 0.308 * - 2.27) = (0.149 * v1) + (0.308 * v2)
0.10579 + (-0.69916) = 0.149 v1 + 0.308v2
−0.59337 = 0.149 v1 + 0.308v2
Dividing both sides by 0.149
v1 + 2.067v2 = −0.59337 - - - - - (2)
From (1)
v2 = 2.98 + v1
v1 + 2.067(2.98 + v1) = −0.59337
v1 + 6.16 + 2.067v1 = −0.59337
3.067v1 = −0.59337 - 6.16
3.067v1 = −6.75337
v1 = −6.75337 / 3.067
v1 = −2.201946 m/s ( to the left)
From v2 = 2.98 + v1
v2 = 2.98 + (-2.201946)
v2 = 0.7780534 m/s ( to the right)
what is a hypothesis reffered to as after being verified by a large number or independent experiments
Answer:
The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.
you walk 6 block east, 2 blocks north, 3 blocks west and then 2 blocks north. the total distance you travel is blocks
Answer:
The answerI travel 13 blocksA person holds a 25 kg (250 newton) bag of cement over his head and moves it a distance of 10 m, taking 2 minutes, while another person carries it on a wheelbarrow that same distance, taking 1 minute.Who does more work ? What is the power of each person?
Explanation:
Assuming the 10 m distance is the vertical displacement, the work done by both people is the same.
Work = force × distance
W = (250 N) (10 m)
W = 2500 J
The power of the first person is:
Power = work / time
P = 2500 J / 120 s
P = 20.83 W
The power of the second person is:
P = 2500 J / 60 s
P = 41.67 W
Find the odd one out a)Planets moving on it's axis,Strings of guitar (being played),Motion of a ferry wheel,The vehicles moving on a straight road. b)Motion of the moon around the earth,Motion of the earth around the sun,Motion of a merry-go-round,Heart-beat in a healthy person. c)Motion of a bullet fired from gun,Motion of a football player in the ground,Motion of a vehicle on a straight road,Motion of an apple falling from a tree
Answer:
a) Strings of guitar (being played)
b) Heart-beat in a healthy person
c) Motion of an apple falling from a tree
Explanation:
a) The motion of the string of a guitar being played is the only motion involving simple harmonic motion.
b) All the other motions are circular motion except the heart beat in a healthy person, which is periodic.
c) The motion of an apply falling from a tree is the only motion under the complete influence of gravity from the onset till the end.
A block is attached to the end of a spring. The block is then displaced from its equilibrium position and released. Subsequently, the block moves back and forth on a frictionless surface without any losses due to friction. Which one of the following statements concerning the total mechanical energy of the block-spring system this situation is true?
1. The total mechanical energy is dependent on the maximum displacement during the motion.
2. The total mechanical energy is at its maximum when the block is at its equilibrium position
3. The total mechanical energy is constant as the block moves back and forth.
4. The total mechanical energy is only dependent on the spring constant and the mass of the block.
Answer:
The correct option is;
3. The total mechanical energy is constant as the block moves back and forth
Explanation:
The total mechanical energy is the sum of the potential and kinetic energies of the system
For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant
For a black and spring system, we have total mechanical energy, E = 1/2×K×A².
Where;
K = Constant
A = The amplitude of motion
Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.
