What the equation given?
x(t)=Acos[tex]\omega[/tex]tMaximum velocity occurs at the equilibrium position
So
x=0Now
x(0)=Acos0[/tex]x(0)=ANow
As we know the formula
[tex]\\ \rm\rightarrowtail V_max=A\omega[/tex]
These expressions can be used
The maximum speed of the oscillating object will be given by [tex]V_{max}=Aw[/tex]
What is oscillation?An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.
The given equation is as follows:
x(t)=Acost
Maximum velocity occurs at the equilibrium position x=0
x(0)=Acos0
x(0)=A
Hence the maximum velocity will be [tex]V_{max}=Aw[/tex]
To know more about Oscillation follow
https://brainly.com/question/26449711
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. Which of the following statements correctly describes the relationship between m1 and m2 and provides evidence from the graphs?
Answer:
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1 describes the displacement
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
What is the graph represents?The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Therefore, M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
Learn more about graph on:
https://brainly.com/question/17267403
#SPJ2
Given the necessary information about the two components of projectile motion, your task is to explain the difference or even describe the two components by their formulas or equations.
The two components of projectile motion include the following:
Horizontal motionVertical motion.What is a Projectile?This is defined as a missile propelled by the application of an external force and allowed to move freely under the influence of gravity and air resistance.
The equation for Horizontal motion Vx = V * cos(α)
Vertical velocity component: Vy = V * sin(α)
Read more about Projectile here https://brainly.com/question/24216590
If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds is given by H = 15t − 1.86t2. (a) Find the velocity (in m/s) of the rock after 1 second. 11.5404 Incorrect: Your answer is incorrect. seenKey 11.28 m/s (b) Find the velocity (in m/s) of the rock when t = a. m/s (c) When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.) t = s (d) With what velocity (in m/s) will the rock hit the surface?
(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.
(b) The velocity of the rock after 2 seconds is 7.56 m/s.
(c) The time for the block to hit the surface is 4.03.
(d) The velocity of the block at the maximum height is 0.
Velocity of the rockThe velocity of the rock is determined as shown below;
Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m
v² = u² - 2gh
where;
g is acceleration due to gravity in mars = 3.72 m/s²v² = (15)² - 2(3.72)(13.14)
v² = 127.23
v = √127.23
v = 11.28 m/s
Velocity of the rock when t = 2 secondv = dh/dt
v = 15 - 3.72t
v(2) = 15 - 3.72(2)
v(2) = 7.56 m/s
Time for the rock to reach maximum heightdh/dt = 0
15 - 3.72t = 0
t = 4.03 s
Velocity of the rock when it hits the surfacev = u - gt
v = 15 - 3.72(4.03)
v = 0
Learn more about velocity at maximum height here: https://brainly.com/question/14638187
An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 39.2 m
Answer: 6067.5 N
Explanation:
Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.
Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.
At an altitude of 1.3x10^7 m above the surface of the earth an incoming meteor mass of 1x10^6 kg has a speed of 6.5x10^3 m/s. What would be the speed just before impact with the surface of earth?Ignore air resistance.
Show all steps.
Answer:
Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.
Explanation:
Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].
Let [tex]m[/tex] denote the mass of the meteor.
Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.
Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.
Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.
The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:
At an altitude of [tex]1.3 \times 10^{7}\; {\rm m}[/tex] about the surface of the earth, the meteor would be approximately [tex]r_{0} \approx 6.371 \times 10^{6}\; {\rm m} + 1.3 \times 10^{7}\; {\rm m} \approx 1.9 \times 10^{7}\; {\rm m}[/tex] away from the surface of planet earth. The meteor would be only [tex]r_{1} \approx 6.371 \times 10^{6}\; {\rm m}[/tex] away from the center of planet earth just before landing.Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.
During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.
Initial [tex]\text{KE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].
Initial [tex]\text{GPE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].
(Note the negative sign in front of the fraction.)
Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:
[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].
[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:
[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].
[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].
Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:
[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].
Substitute in the values and evaluate:
[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].
(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.
A source of light emits photons with a wavelength of 8.1 x 10-8 meters. What is the frequency of this light
Answer:
Explanation:
Speed of light v = 3 x 10⁸ m/s
wavelength λ = 8.1 x 10⁻⁸ m
frquency f = v/λ = 3.7 x 10¹⁵ Hz
If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,
The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸
=3.7 × 10¹⁵ Hz
Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz
To learn more about wavelength, refer to the link given below ;
brainly.com/question/7143261
#SPJ5
A women runs a kilometer using a of 210N and a pwower output of 500W how long does it take this woman to complete 1 kilometer
Answer:
7 minutes
Explanation:
Explanation:
Power = work / time
Power = force × distance / time
500 W = 210 N × 1000 m / t
t = 420 s
t = 7 min
Why is it more difficult to roller blade across rough asphalt than it is to roller blade across smooth concrete?
A. Asphalt has more mass than concrete.
B. Friction on the roller blades is greater on asphalt than on concrete.
C. Roller blades exert a greater force on asphalt than they do on concrete.
D. The inertia of the roller blades is greater on asphalt then on concrete.
Answer:
The answer is B. Friction on the roller blades is greater on asphalt than on concrete.
Have an amazing day!!
Please rate and mark brainliest!!
A basketball player must jump 1.2m in the air in order to dunk. What is the total time he is in the air?
Answer:
0.98 [sec].
Explanation:
1) the time when the player is moving up equals the time the player falls, it can be 't'. Then
2) the equation of moving down can be written as
[tex]\frac{g*t^2}{2} =h, \ where[/tex]
g=10 [m/s²], t - time, h - the given height.
3) according to the formula above the time spent to moving down is:
[tex]t=\sqrt{\frac{2h}{g}};[/tex]
[tex]t=\sqrt{\frac{2.4}{10}}=\sqrt{0.24} =0.49[s].[/tex]
4) finally, the total time is t*2=0.98[sec].
The weight of your science book presses on the desk.
A. Gravity pulls on the book.
B. Gravity pulls on the desk.
C. The desk pushes on the book.
Answer:
A. Gravity pulls on the book :)
Have an amazing day!!
Please rate and mark brainliest!!
Topic : Collisions and Conservation of Momentum
1. Two children are standing still on ice and push off each other. Each has a momentum after the push. Do they violate the conservation of law? Initial momentum is zero.
2. Describe how the conservation law works with elastic and inelastic collisions.
3. In general, for elastic collisions with stationary objects, when will the first object bounce back? When will it follows the stationary object after the collision?
4. What is the use of airbags in vehicles?
5. Why are seatbelt designed to be stretchable?
It is amount of force or motion that cause the movement or moving of body.
Thus it is well explained.
To learn more about the collisions refer to link ;
https://brainly.com/question/7694106
1.No they do not violate the conservation of law. Because the initial momentum can be zero.
2. The momentum is conserved in both elastic and inelastic collision.
What is the law of conservation of momentum?The momentum of the body before the contact is always equal to the momentum of the body after the impact, according to the rule of conservation of momentum.
According to the law of conservation of momentum;
Momentum before collision =Momentum after collision
[tex]\rm m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]
1.No they do not violate the conservation of law. Because the initial momentum can be zero.
2. The momentum is conserved in both elastic and inelastic collision. Hence the conservation law works with elastic and inelastic collisions.
3. An elastic collision is one in which kinetic energy is conserved, meaning that the total kinetic energy before and after the impact is the same.
The object bounces once after the body gets struck. When the body's final velocity is zero it follows the stationary object after the collision.
4.. Airbags are meant to inflate automatically in the case of an abrupt deceleration or impact force indicative of a collision.
When the bag is fully inflated, it protects the driver and/or passenger by Increasing cushioning around the driver's neck, head, and spine.
5. Stretchable seatbelts are meant to allow a passenger's considerable velocity to be slowly reduced, preventing the passenger from being flung forward abruptly.
As a result, if the automobile collides, serious casualties can be avoided.
Hence no they do not violate the conservation of law. Because the initial momentum can be zero.
To learn more about the law of conservation of momentum refer;
https://brainly.com/question/1113396
Who is the father of the humans?
Answer:
Adam (Arabic:, romanized:dam) is said to be the first human person on Earth, as well as the first prophet (Arabic:, nab) of Islam. Muslims see Adam's role as the father of the human race with veneration.
Explanation:
Answer:
Adam is the name given to the first person in Genesis 1-5. Aside from being the name of the first man, Adam is also used as a pronoun in the Bible, both individually as "a human" and collectively as "mankind."
Explanation:
Why is a strong skeletal system important for leading an active lifestyle?
Answer:
Explanation:
Because if you don't have strong bones, you will have problems with your health. So, a strong skeletal system is important for leading a healthy life.
A crate of oranges is shoved across a grocery store floor with a force of 100 N for a distance of 1 meter. The crate travels an additional 1 meter after the shove and stops just short of the display. You can conclude: the magnitude of work done by frictional forces on the crate is 100 J the magnitude of work done by the applied force on the crate is 100 J the magnitude of the force of friction on the crate is 50 N all of the above conclusions are correct none of the above conclusions are correct
The statements that are true are;
the magnitude of work done by frictional forces on the crate is 100 Jthe magnitude of work done by the applied force on the crate is 100 JWhat is work done?Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.
Hence, the following are true;
the magnitude of work done by frictional forces on the crate is 100 Jthe magnitude of work done by the applied force on the crate is 100 JLearn more about work: https://brainly.com/question/18094932?
17) Which object would likely have the greatest velocity
a. a bouncy ball
b. a bowling call
a go-kart
d. a school bus
Answer: b or d
Explanation: b or d
Answer:
a
Explanation:
F= ma
interestingly
when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases
(man, PHYSICS IS JUST THE BESY)
A: a
object with the smallest mass has largest acceleration
What is the name of an ionic compound that consists of lithium ion LI and iodide ion (I)?
Answer:
Lithium iodide
Explanation:
it's formula is LiI
Lithium iodide is a chemical compound composed of lithium and iodine, which when exposed to air turns yellow due to the oxidation of iodide to iodine. It has a chemical formula LiI.
The products, lithium iodide (LiI), sodium iodide (NaI), and cesium iodide (CsI) look like typical ionic compounds; they are all white crystalline solids
a hunter shoots a stone from his catapult with an initial velocity of 40 m per seconds add elevation of 60 degree with the aim of hitting a bird 145n away.
1. calculate the time of flight
Answer:
approximately 7 seconds
Explanation:
The given scenario is just an example of Projectile Motion.
In this question, we are given:
- Initial Velocity (u) = 40 m/s
- Angle of Elevation (θ) = 60 degrees
- Distance of Bird (s) = 145 meters [Not used for the time of flight]
Time of Flight:
[tex]T = \frac{2uSin(\theta)}{g}[/tex]
we will use g = 9.8 m/s/s since we are not told otherwise.
[tex]T = \frac{2(40)Sin(60 degrees)}{9.8}[/tex]
[tex]T = \frac{40\sqrt{3}}{9.8}[/tex]
T = 4.081(√3) seconds
rounded to 7 seconds
Mention the objective of the Experiment?
Answer:
I don't understand your question ❓,the object.....of what experiment
There are two space ships traveling next to each other. The first one is 500
Kg and the second one is 498 Kg. Since they are 35 meters apart, what is
the force of gravity between the two space ships?
This question involves the concept of Newton's law of gravitation.
The force of gravity between the two spaceships is "1355.78 N".
Newton's Law Of GravitationAccording to Newton's Law of Gravitation:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
where,
F = force of gravity between ships = ?G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²m₁ = mass of first ship = 500 kgm₂ = mass of second ship = 498 kgr = distance between ships = 35 mTherefore,
[tex]F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(500\ kg)(498\ kg)}{(35\ m)^2}\\\\[/tex]
F = 1355.78 N = 1.356 KN
Learn more about Newton's Law of Gravitation here:
https://brainly.com/question/9373839
Two space ships traveling next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.
It is given that the First spaceship's weight ([tex]m_{1}[/tex]) is 500 kg,
The second spaceship's weight ([tex]\rm m_{2}[/tex]) is 498 kg.
The distance between spaceships (r) is 35 meters.
It is required to find the Force of gravity between these spaceships.
What is Gravitational force?It is defined as the force which attracts any two masses in the universe.
By Newton's law of Gravitation:
[tex]\rm F= \frac{Gm_1m_2}{r^2}[/tex] , Where
[tex]\rm F = The\ force \ of \ gravity \ between \ the \ spaceships\\\rm G= Universal\ Gravitational \ Constant = 6.67 \times 10^{-11} N.m^2/kg^2[/tex]
Putting values in the above formula:
[tex]\rm F = \frac{(6.67\times 10^{-11} N.m^2/kg^2)(500kg)(498kg)}{(35m)^2}[/tex]
F = 1355.78 N = 1.356 KN
Thus, Two spaceships travel next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.
Learn more about Gravitational Force here:
https://brainly.com/question/3009841
A solenoid has 450 loops each of radius 0.0254 m. The field increases from 0 T to 3.00 T in 1.55 s. What is the EMF generated in the coil? (Hint: What is the area of a circle) (Unit = Volts)
Answer:
0.175 second
Explanation:
i hope it helps
I’m sorry, but the person you are calling has a voice mailbox that has not been set up yet.
Please try again later.
Bye Bye!
Answer:
bye bye
Explanation:
Answer: Or please call again to see if they reached the phone yet.
Explanation: It always stops ringing right before someone gets to there phone so call again in 2 minutes. :3
A stage has a radius R = 2.00 m and a mass M = 100 kg. The stage, which has a circular disk
shape, rotates in a horizontal plane, without any friction, about a vertical axle. Sandra has a
mass of m = 60.0 kg and she walks from the rim of the stage toward the center. Calculate the
angular speed when she reaches a point r = 0.500 m from the center, if the angular speed of the
system is 2.0 rad/s when Sandra is at the rim
i want the answer of this exercise
What is the approximate electrostatic force between two protons separated by a distance of 1.0 x 10-6 meter?
1) 2.3 x 10-16 N and repulsive
2) 9.0 x 1021 N and repulsive
3) 9.0 x 1021 N and attractive
4) 2.3 10-16 N and attractive
Answer:
2.3×10^-16
Explanation:
Charge of proton = 1.6×10^-19
let q(a) and q(b) be the two protons.
According to the Coulumbs Laws ;
F = kq(a)×q(b) / r^2
where k = 9.0×10^9 and r is the distance of seperation.
So having a distance of 1.0×10^-6 in our case ;
F = (9.0×10^9×1.6×10^-19×1.6×10^-19)/(1.0××10^-6)^2
F = (2.304×10^-28)/(1.0×10^-12)
F = 2.304×10^-16N
Since the both charges are positively charged they will repel .
Express this as y=mx + c
1.
[tex] \frac{1}{y} = \alpha ( {x - 6)}^{ - b} [/tex]
Answer:
let alpha be y,
1/2=y(x-6)^(-b)
1/2=y(1/x-6)
1/2=y/x-6
x-6=2y
x-6/2=y
(1/2)x-(6/2)=y
y=(1/2)x-3
PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!
An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100-N weight attached, the pointer indicates "40" on the scale as shown. Using a 200-N weight instead results in "60" on the scale. Using an unknown weight X instead results in "30" on the scale. The weight of X is:
Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L) (1)
200 = k(60 - L) (2)
(2)/(1): 2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
Where is the electric field between a charged balloon and a charged piece of paper 0.04 m away?
A. The electric field is in the region around the balloon and paper where there are outside charges that affect them.
B. The electric field is in the 0.04 m distance between the balloon and paper and not anywhere else around either the paper or the balloon.
C. The electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
D. The electric field is in the region in which the forces come in contact.
The electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
What is electric field?An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.
Thus, for the charged balloon and a charged piece of paper 0.04 m away, we can conclude that, the electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
Learn more about electric field here: https://brainly.com/question/14372859
#SPJ1
Organic farming follows the
principles of health and
A. economic gain
B. wealth
C. profits
D. ecology
Answer:The Answer is economic gain
Explanation:
Q. 1 MWH is equal to ------- joules
a.3.6*10^10
b.3.6*10^6
c.3.6*10^9
d.3.6
can somebody please help
The amplitude of the wave on the given sinusoidal wave graph is 10 cm.
What is amplitude of wave?
The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.
Amplitude of the wave is calculated as follows;
From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.
Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.
Learn more about wave amplitude here: https://brainly.com/question/25699025