The correct option is b. The one with the lowest mass.
An object's kinetic energy is determined by
k=1/2mv^2
where
m is the object's mass.
v is the object's speed.
The three missiles in this puzzle have varying masses but the same beginning kinetic energy.
The three projectiles will all have the same kinetic energy when they hit the ground because mechanical energy is conserved, assuming there is no air resistance (because the potential energy that they have lost is the same, since they have been launched from the same height, and they reach the same final altitude, the ground).
hence,
K1=k2=k3
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A rocket is launched from the surface of the earth with a speed of 9.0x103 m/s. What is the maximum altitude reached by the rocket? (MEarth=5.98x1024 kg, REarth=6.37x106 m)
From the Law of conservation of energy, we know that the sum of the kinetic and potential energy of the rocket is the same at the surface of the Earth and at the maximum altitude. Nevertheless, the kinetic energy of the rocket when it is at the maximum altitude is 0:
[tex]\begin{gathered} K_1+U_1=K_2+U_2 \\ K_2=0 \\ \Rightarrow K_1+U_1=U_2 \end{gathered}[/tex]The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]On the other hand, the gravitational potential energy for big changes in altitude (comparable to the radius of the Earth) is given by the expression:
[tex]U=-\frac{GMm}{r}[/tex]Where M is the mass of the Earth, m is the mass of the rocket, r is the distance from the center of the Earth to the rocket and G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\cdot\frac{m^2}{\operatorname{kg}}[/tex]At the beggining of the movement, the value of r corresponds to the radius of the Earth:
[tex]U_1=-\frac{GMm}{R_E}[/tex]At the end of the movement, the value of r corresponds to the radius of the Earth plus the maximum altitude h:
[tex]U_2=-\frac{GMm}{R_E+h_{}}[/tex]Substitute the expressions for U_1, K_1 and U_2 and simplify the equation by eliminating the factor m:
[tex]\begin{gathered} \frac{1}{2}mv^2-\frac{GMm}{R_E}=-\frac{GMm}{R_E+h} \\ \Rightarrow\frac{1}{2}v^2-\frac{GM}{R_E}=-\frac{GM}{R_E+h} \end{gathered}[/tex]Isolate the term GM/(R_E+h):
[tex]\Rightarrow\frac{GM}{R_E+h}=\frac{GM}{R_E_{}}-\frac{1}{2}v^2[/tex]Divide both sides by the factor GM:
[tex]\Rightarrow\frac{1}{R_E+h}=\frac{1}{R_E}-\frac{v^2}{2GM}[/tex]Take the reciprocal to both sides of the equation:
[tex]\Rightarrow R_E+h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}[/tex]Isolate h:
[tex]h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}-R_E[/tex]Substitute the values of each variable: R_E=6.37x10^6m, M=5.98x10^24kg, G=6.67x10^-11 N*m^2/kg^2, and v=9.0x10^-3 m/s:
[tex]\begin{gathered} h=\frac{1}{\frac{1}{6.37\times10^6m}-\frac{(9.0\times10^3\cdot\frac{m}{s})^2}{2(6.67\times10^{-11}N\cdot\frac{m^2}{kg^2})(5.98\times10^{24}kg)}}-6.37\times10^6m \\ =18.03\times10^6m-6.37\times10^6m \\ =11.7\times10^6m \end{gathered}[/tex]Therefore, the maximum altitude reached by a rocket with an initial speed of 9.0x10^3m is:
[tex]11.7\times10^6m[/tex]A piece of rock weighs 5 Newtons. A force F is applied to it and it produces an acceleration of a. Now, if there is a second piece of rock, what force needs to be applied to the second piece of rock to produce an acceleration of 8a?
Given,
weight of a piece of rock, W = mg = 5N
So, mass of a piece of rock is m = W ÷ g = 5N ÷ 9.8 m/[tex]s^{2}[/tex] = 0.51 kg.
When a force F is applied, it produces an acceleration 'a'. From Newton's second law of motion, F = ma = 0.51 a.
⇒a = F / 0.51.
Acceleration produced by 2nd piece of the rock is a' = 8a.
Assuming the mass of 2nd piece of rock to be same as the 1st piece, force to be applied on 2nd piece is F' = m'a'.
⇒ F' = m' × 8a = 8m'a
If the mass of both pieces of rock are equal, then m' = m = 0.51 kg.
⇒ F' = 0.51kg × 8a
⇒ F' = 0.51kg × 8 (F / 0.51)
⇒ F' = 8F.
Thus, a force equal to 8 times the force applied on 1st piece should be applied on the 2nd piece of the rock to produce an acceleration of 8a (if their masses are equal.
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024 (part 1 of 3) 10.0 points
A 4.0 kg block is pushed 1.0 m at a constant
velocity up a vertical wall by a constant force
applied at an angle of 30.0
◦ with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s
2
.
1 m
30◦
4 kg
F
Drawing not to scale.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.
The work done on the force on the block is 23.516J
We are given that,
The mass of the block = m= 4.0 kg
The vertical component of the wall at an angle = sinθ = sin 30°
The horizontal component of the wall at an angle = cosθ = cos30°
The acceleration due to gravity = g = 9.81 m/s²
The coefficient of kinetic friction between the block and the wall= μ=0.40
So that to calculate the work done by the force on the block can be given by the expression, for vertical and horizontal component
Wf = Fycosθ
The Wf is the work done by the force Fy (on y-axis) on the block by the horizontal component,
Wf = Fxsinθ
The Wf is the work done by the force Fx (on x-axis) on the block by the vertical component,
Net Fy = sinθ N -mg =0
Net Fx = 0
Therefore, the work done by the force on the block can be assumed by above equations,
Wf = (mgdsinθ)/(sinθ - μcosθ)
Wf = (4.0kg)(1.0m)(9.81m/s²)/sin30° -(0.40)cos30°
Wf = 33.981/1.445 J
Wf = 23.516J
Thus , the work done by the force on the block would be 23.516J
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jumping up before the elevator hits. after the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 m. during the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 5.0 ms. (assume that neither the passenger nor the cab rebounds.) what are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? if the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab hits the bottom of the shaft,
a) The magnitude of the impulse is 2.39 × [tex]10^{3}[/tex] N. b) Average force on the passenger during the collision is 4.78 × [tex]10^{5}[/tex] N. c) impulse is 1.6 × [tex]10^{3}[/tex] N.s d) The corresponding average force would be 3.52 × [tex]10^{5}[/tex] N.
a) By energy conservation, the speed of the passenger when the elevator hits the floor is
1/2 mv² = mgh
v = √2gh = [tex]\sqrt{2(9.8)(36)}[/tex]
= 26.6 m/s
The magnitude of the impulse is
J = |Δp| = m|Δv| = mv = (90kg) (26.6m/s) ≈ 2.39 × 10³ N.s
b) With duration of Δt = 5.0 × [tex]10^{-3}[/tex] s for the collision, the average force is
[tex]F_{avg}[/tex] = J/ Δt = 2.39×10³ N.s / 5.0 × [tex]10^{-3}[/tex] s
≈ 4.78 × [tex]10^{5}[/tex] N
c) If the passenger were to jump upward with a speed of [tex]v^{'}[/tex]= 7.0 m/s, then the resulting downward velocity would be
[tex]v^{n}[/tex] = v- [tex]v^{'}[/tex]
= 26.6m/s - 7.0 m/s
= 19.6 m/s
The magnitude of the impulse becomes
[tex]J^{n}[/tex] = |Δ[tex]p^{n}[/tex]| = m|Δ[tex]v^{n}[/tex]| = m[tex]v^{n}[/tex] = (90kg) (19.6 m/s)
≈ 1.76 × 10³³N.s
d) The corresponding average force which is
[tex]F_{avg } ^{n}[/tex] = [tex]J^{n}[/tex]/ Δt = 1.76 × 10³ N.s/ 5.0 × [tex]10^{-3}[/tex] s
≈ 3.52 × [tex]10^{5}[/tex]N
Therefore the magnitude of the impulse is 26.6 m/s, average force on the passenger during the collision is 4.78 × [tex]10^{5}[/tex]N, the magnitude of the impulse is 1.76×[tex]10^{3}[/tex]N.s, average force is 3.52 × [tex]10^{5}[/tex].
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light enters glass from air. the angle of refraction will be: group of answer choices less than the angle of incidence. equal to the angle of incidence. greater than the angle of incidence
Less than the angle of incidence, the angle of refraction will exist.
What is angle of refraction?Angle of refraction refers to the angle formed by the refracted ray and the normal at the point of incident.
In a mirror, the angle of incidence and angle of reflection are the same. If the incident ray falls along the normal, the angle of incidence for a plane mirror is 0 degrees rather than 90 degrees.
The behavior of light is described by Fermat's principle, which states that a light ray always chooses the shortest route to its destination. On reflection, the exact same behavior is seen. The angle of incidence and the angle of reflection are equal when a ray is reflected from a plane.
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does increasing the magnitude of a uniform magnetic field through which a charge is traveling necessarily mean increasing the magnetic force on the charge? does changing the direction of the field necessarily mean a change in the force on the charge?
Since the magnetic force and velocity are orthogonal, the velocity only changes in direction but not in magnitude . Circular motion is the end consequence. The force is in the opposite direction from the right-hand rule and has a negative charge.
Uniform In circular motion, an object moves in two dimensions with a constant speed in a fixed circular direction; however, because the object's direction changes at every point, the velocity also changes, and the direction at each point is always the tangent.
In physics, magnetic force is referred to as an object's maximal size and direction. Both vector and scalar values use magnetic force as a common factor. We understand that by definition
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Answer: The velocity simply changes in direction, not magnitude, as the magnetic force and velocity are orthogonal.
Explanation: The velocity simply changes in direction, not magnitude, as the magnetic force and velocity are orthogonal. The resultant motion is circular. The force has a negative charge and is directed in the opposite direction from the right-hand rule.
Uniform In a circular motion, an item moves in two dimensions with a constant speed in a fixed circular direction; yet, because the object's direction varies at each point, the velocity also alters, and the direction at each point is consistently the tangent.
As an object's maximum size and direction, the magnetic force is referred to in physics. Magnetic force is a factor that is present in both scalar and vector values. We comprehend that by nature.
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when the pendulum bob is at its maximum height (farthest from the center of the earth), is the gravitational potential energy of the bob-earth system at a maximum, minimum or some value in between? briefly explain.
Both gravitational potential energy (PE) and kinetic energy (KE) are present on a swinging pendulum bob. The KE is at its highest and the PE is zero when the bob is in its lowest position. The KE is zero and the PE is at its highest value when the bob is in its highest position on both sides.
When does the pendulum's speed reach its highest point?The pendulum moves at its fastest while it is passing through the middle position. We could interpret this in terms of energy conservation. Bob is at his shortest height when he is in the average position. When Bob is positioned at an extreme, such as the right, it is slightly higher than when it is in the middle.
What happens to a pendulum's length when it is full?The bob's center of mass serves as the standard for determining the pendulum's "length." The center of gravity of the bob would fluctuate in a convoluted manner as the mercury drained out of a simple spherical bob with a small hole in the bottom. The center of gravity would be in the middle of the bob when it is totally full.
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a swan on a lake gets airborne by flapping its wings and running on top of the water. (a) if the swan must reach a velocity of 5.80 m/s to take off and it accelerates from rest at an average rate of 0.305 m/s2, how far (in m) will it travel before becoming airborne?
The swan will travel 55.15 m during uniform motion before becoming airbone.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From the question above, we know that
vo = 0 m/s
vt = 5.8 m/s
a = 0.305 m/s²
By using the second equation, we can calculate the distance before becoming airbone
vt² = vo² + 2a . s
5.8² = 0² + 2 . 0.305 . s
0.61s = 33.64
s = 55.15 m
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To prevent water pollution, a factory proposes pumping its wastes into the ground instead of into a river. Would you support this proposal? Why or why not?
Select the correct answer. when a small star dies, which of these celestial objects is it most likely to help create? a. a planet b. a star c. a moon d. a meteor e. a comet
It is best to choose option B. When a star runs out of fuel, it can no longer undergo the merger process and dies. The primary elements of stars are helium and hydrogen.
A white dwarf star's maximum mass is determined by Chandrasekhar's limit, a sort of maximum restricted value. The Chandrasekhar limit is often valued at 1.44 times the mass of the sun, or around 2.765 1030 kg, on a global scale.
In other words, the Chandrasekhar limit is the highest figure that may possibly reach 1.44 times the mass of the sun.
When a high-mass star runs out of gas to burn, it enlarges and changes into a red supergiant. While the bulk of stars gradually fade away, the supergiant stars, often known as supernovae, demolish themselves in a massive explosion. New stars can develop as a result of massive stellar deaths.
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assuming that the air friction is responsible for up to 15% momentum or energy loss, discuss how your results have (roughly) confirmed or contradicted the conservation law for elastic collisions if there is no air friction. do different mass configurations show the same physics of conservation (or non-conservation)? why or why not? g
Due to air friction different mass configurations show the same physics of non-conservation.
Momentum in an easy way is an amount of motion. right here quantity is measurable because if an item is transferring and has mass, then it has momentum. If an item no longer passes then it has no momentum. but, in normal lifestyles, it has significance but many humans did not understand it.
Momentum may be described as "mass in motion. All items have mass; so if an item is transferring, then it has momentum - it has its mass in motion. the quantity of momentum that an item has relies upon two variables.
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145-g baseball is dropped from a tree 12.0 m above the ground. (a) with what speed would it hit the ground if air resistance could be ignored? (b) if it actually hits the ground with a speed of what is the average force of air resistance exerted on it?
The required speed is 15.35 m/s, if air resistance is ignored.
The magnitude of average force of air resistance is 1.0341N.
a) V = √2gh
The required speed is
V = √2(9.8)(12)
V = 15.35m/s
b) W = 1/2m([tex]v^{2}[/tex]-[tex]u^{2}[/tex])
W = (mg - f)h
( (0.145)(9.8) - f)12 = 1/2(0.145)(([tex]8^{2}[/tex]))
(1.421 - f)12 = 4.64
Force of an air resistance = 1.0341N
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a fish looking straight up toward the smooth surface of a pond receives a cone of rays and sees a circle of light filled with the images of sky and birds and whatever else is up there. this bright circular field is surrounded by darkness. explain what is happening and compute the cone angle.
The phenomenon that occurs here is known as the snell's window, and the cone angle is 98 degrees.
To calculate the cone angle,
The cone angle is equal to two times the critical angle which is say, Ф
To calculate the critical angle Ф take the equation sinФ = n1÷n2 where
n1 is the refractive index of air, and n2 is the refractive index of water.
The refractive index of air, n1 = 1,
The refractive index of water, n2 = 1.33,
Substituting these values in the equation sinФ = n1÷n2, the equation becomes
sinФ = 1÷1.33
solving for Ф
Ф = [tex]sin^{-1}[/tex](1÷1.33)
Ф = 49°
Cone angle = 2×Ф
Cone angle = 2×49
Cone angle = 98°.
The phenomenon taking place here is known as snell's window. In this phenomenon, the person under the water sees everything above the surface through a cone of light whose width is about 96°-98°. Whatever is outside this cone appears dark to the person who is under the water. This is caused due to the refraction of light in different mediums.
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a 5.25-g bullet is fired into a 360-g block of wood suspended as a ballistic pendulum. the combined mass swings up to a height of 8.75 cm. what was the magnitude of the momentum of the combined mass immediately after the collision?
magnitude of the momentum of the combined mass immediately after the collision p = 0. 478 (kg. m/sec)
Mass of bullet : m = 5. 25 (g) = 0. 00525 (kg)
Mass of block : M = 360 (g) = 0. 360 (kg)
Height : h = 8. 75 (m) = 0. 0875 (m)
Here we need to find the momentum :
Now,
As we know that,
The conservation of energy :
Now,
Kinetic energy = Potential energy
= 1/2( m + M)v² = (m+M)gh (m,M is mass of bullet and block)
Now,
v = [tex]\sqrt{2gh}[/tex]
Now,
Momentum can be defined as :
Here,
p = (m+ M) [tex]\sqrt{2gh}[/tex]. . . eqn 1
Putting the values in eqn 1 :
We get,
p = 0.478 (kg. m/sec)
Hence, the momentum : p = 0. 478 (kg. m/sec)
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use the values from practice it to help you work this exercise. what if an additional mass is attached to the ball? how large must this mass be to increase the downward acceleration by 35%?
If the additional mass is attached to the ball, it must be large enough to increase the downward acceleration by 35%.
What is acceleration?
An object is said to have been accelerated if its velocity changes. An object's velocity can change depending on whether it moves faster or slower or in a different direction. A falling apple, the moon orbiting the earth, or a car stopped at a stop sign are a few instances of acceleration. Through these illustrations, we can see that acceleration happens whenever a moving object changes its direction or speed, or both. Since acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is either the first derivative of velocity in relation to time or even the second derivative of position in relation to time.
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select the best answer. the logarithmic decrement is: a. the ratio of the damping in the system to the critical damping b. the ratio of the successive amplitudes in a time series c. the ratio of the damping in one system compared to another system d. the total damping present in the system
Logarithmic decrement is the ratio of successive amplitudes in a time series.
The logarithmic decrement δ is equal to the natural logarithm of the ratio of two successive maximum deflections x of oscillating quantity in the same direction:
δ = log( [tex]x_{1} /x_{2}[/tex])
Logarithmic decrement is used to find the damping ratio of an underdamped system in the time domain. It is defined as the natural logarithm of the ratio of any two successive amplitude on the same side of the mean line.
Therefore we get that logarithmic decrement is the ratio of successive amplitudes in a time series.
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A substance added to gasoline is found in the wells of people who live miles from a gas station. Is this an example of a point source of pollution or a nonpoint source of pollution? Explain.
I need to get it done right now
The substance added to gasoline and found in the wells of people who live miles from a gas station is an example of non-point source of pollution.
What are point source pollution?Point source pollution are sources of pollution, radiation, waves, fluid or other substance that has one specific location.
On the other hand, non-point source pollution is a combination of pollutants from a large area rather than from specific identifiable sources such as discharge pipes.
According to this question, a substance added to gasoline is found in the wells of people who live miles from a gas station. This suggests that the pollutants enter from a wide array of sources.
Therefore, it can be said that the source of pollution is a non-point source.
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a machine has a efficiency of 20%, find the input work if the output work is 140 J
The input work of a machine with an efficiency of 20% and an output work of 140 J is 28 J
What is input work of a machine?
Input work is the work done on a machine as the input force acts through the input distance.
To calculate the input work of the machine, we use the formula below.
Formula:
E(%) = (Wi/Wo)×100......... Equation 1Where:
E(%) = Efficiency of the machineWi = Input workWo = Output work.Make Wi the subject of the equation
Wi = E(%)×Wo/100....... Equation 2From the question,
Given:
E(%) = 20%Wo = 140 JSubstitute these values into equation 2
Wi = 20×140/100Wi = 28 JHence, the input work of the machine is 28 J.
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A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is 0. 35. What horizontal force is required for an acceleration of 2. 0 m/s2?
The horizontal force required for an acceleration of 2.0 m/s² is equal to 54.3 N.
What is the force of kinetic friction?The force of kinetic friction is a type of frictions that acts on a moving object on a surface. In order to calculate the force of kinetic friction, we use the formula F = μN where F is the force of kinetic friction, μ is the coefficient of the kinetic friction, and N is the normal force. However, we need to take into account the force along the x-axis. Therefore, the formula ∑F = μN + ma will help us find the horizontal force.
Let ∑F represent the net force, N represent the normal force, m represent the mass of the box, g represent the acceleration of gravity, µ represent the kinetic friction, and a represent an acceleration of 2 m/s².
∑F = µN + ma
∑F = µmg + ma
∑F = m (µg + a)
∑F = 10 kg (0.35 × 9.8 m/s² + 2 m/s²)
∑F = 10 kg (3.43 m/s² + 2 m/s²)
∑F = 10 kg (5.43 m/s²)
∑F = 54.3 kg × m/s²
∑F = 54.3 N
The net force is equal to 54.3 N. That is the horizontal force required!
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A cart is set up as shown below, with three fans directed to the left and two of the fans running. The motion of the cart is represented by the v vs t graph shown. If the experiment were repeated with all three fans running, what might the resulting v vs t graph look like?
A. Graph A
B. Graph B
C. Graph C
D. Graph D
A ball is dropped from a height of 23 meters. Find the velocity of the ball just prior to
striking the ground.
round to nearest hundreds
21.168 m/s velocity of the ball just prior to striking the ground.
Round to nearest hundreds is 21.17
What is the meaning of velocity?The direction of the movement of the body or the object is defined by its velocity. Most of the time, speed is a scalar quantity. In its purest form, velocity is a vector quantity. It measures how quickly a distance changes. It is the rate at which displacement is changing.
“equations of motion”
How long does it take?
s = ut + ½ at²
23 m = 0 + ½ *9.81m/s² * t²
23 m = 4.905 t²
t² = 23 / 4.905
t² = 4.689
t =2.165s
What velocity will it hit the ground?
Take ↓ direction as +
v = u + at
v = 0 + 9.81 m/s² * 2.16 s
v =21.168 m/s
Velocity = 21.168 m/s downwards. ( have left you to decide on significant digits in the answer.
round to nearest hundreds is 21.17
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a certain amount of work w is required to accelerate a motorcycle from rest to a speed v. how much work is required to accelerate the same motorcycle from rest up to a speed v/4?
The work required to move the motorcycle from rest to a speed of v/4 is 1/16 times of the initial work w.
As we already know, from the work energy theorem,
That work done by all the conservative forces and all the none conservative forces on the body is equal to the change in kinetic energy of that body.
So, in formula form, it can be written as,
W = ∆K
Where w is work done on the body and ∆K is the change in kinetic energy of the body.
So, if now we assume, that the work done to increase the velocity of the body from zero to V is w, then we can write,
w = 1/2mv² ..... Equation 1
m is the mass of the body.
Now, as per the question,
Let us say the work done needed to increase the speed of the body from rest to v/4 is w', then we can write,
w' = 1/2m(v/4)² ..... Equation 2
Now,
Dividing equation 2 by 1,
We get,
w'/w = 1/16
So, the value of w' is w/16.
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A manufacturer of echo effects for music recording studios has built a prototype of an echo device that sounds exactly like a beloved echo machine from the 1970s. even though its computer chips are cheap, they are hard to find because the compounds originally used to make the chips are no longer being manufactured. what type of issue does the effects manufacturer need to overcome in order to be able to mass produce this echo device? availability cost effectiveness safety time effectiveness
Since sound expands and loses intensity as it moves farther away, the echo is weaker than the initial sound and has a smaller amplitude. For instance, when a speaker's voice is heard and its echo is compared.
One kind of audio effect that involves gradually delaying a signal is an echo effect. In this instance, after some amount of time, listeners hear the same signal repeated audibly. When the time delay is relatively long (more than 30 milliseconds), listeners can distinguish between the echoes.
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Answer:
D: Availability
Explanation:
correct on edge 2023
a small sphere is launched horizontally with a speed v0 toward a target a horizontal distance d away. the sphere lands on the target that is at height h below the height from which the sphere was launched. if the sphere were launched again, but with a speed 2v0 , how high above the target would the sphere have to be launched horizontally to hit the same target, still a horizontal distance d away? responses
The sphere has to be launched 4 times the initial height, h, to hit the same target.
What is the time of motion of the first sphere?The time taken for the first sphere to hit the target is calculated as follows;
d = V₀t₁
where;
V₀ is the initial horizontal velocityd is the horizontal distance travelled by the spheret is the time of motiont₁ = d/V₀
The height travelled by the ball;
h = ¹/₂gt₁²
t₁² = 2h/g
t₁ = √(2h/g)
The time of motion of the second sphere when it is launched at a speed of 2V₀.
t₂ = d/2V₀
t₂ = ¹/₂(d/V₀)
t₂ = ¹/₂(t₁)
The height of fall of the second sphere;
¹/₂(t₁) = √(2h/g)
[¹/₂(t₁)]² = 2h/g
t₁²/4 = 2h/g
t₁² = 4(2h/g)
Thus, the height the sphere need to be launched in order to hit the same target is determined by applying kinematic equation.
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a 90.0-kg man is standing on a scale in an elevator that is accelerating upward at 2.00 m/s2. what will be the scale reading? a. 696 n ob. 788 n oc947 n od. 883 n oe. 1,060
The scale reading of a man standing on a scale in an elevator that is accelerating upward at 2.00 m/s2 is 788 N.
Acceleration is the price of change in the velocity of an object with the recognition of time. Accelerations are vector portions. The orientation of an item's acceleration is given by way of the orientation of the internet pressure acting on that item.
Acceleration is the price at which pace adjustments with time, in phrases of both speed and course. A factor or an object shifting in an immediate line is elevated if it hurries up or slows down. movement on a circle is expanded despite the fact that the rate is constant, due to the fact the path is constantly changing.
f = mg - ma
= 900 - 180
= 788 N
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A hiker, of mass 80kg, walks up a mountain, 700m above sea level, to spend the night at the top in the first overnight hut. The second day she walks to the second overnight hut, 400m above sea level. The third day she returns to her starting point, 200m above sea level. What is the potential energy of the hiker at the first hut (relative to sea level)?
The potential energy of the hiker at the first hut is mathematically given as
v = 548800J
What is the potential energy of the hiker at the first hut?Generally, the equation for potential energy is mathematically given as
v = mgh
where,
m = 80kg, g = 9.8, h = 700
therefore
v = 80 x 9.8 x 700
In conclusion, the potential energy of the hiker at the first hut
v = 548800J
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What is the force of gravity pulling on a 12kg block accelerating at 10m/s²?
120N is the force of gravity pulling on a 12kg block accelerating at 10m/s²
F = m × a
F = 10 ×12
F= 120N
How is the potential energy of gravity stored?The mass (m) and height (h) of the item determine how much gravitational potential energy is there. Based on an object's high position relative to a lower location, gravitational energy is potential energy that is stored in the object.
This energy can be utilised later to move an item since it can be stored and utilised at a later time. Three variables—mass, gravity, and height—determine the gravitational potential energy. Energy is directly inversely proportional to all three variables.
Because the highest amount of energy that may be converted into other forms is called potential energy. You need more energy to move the thing at a greater height.
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Two litres of water, initially at 20 °C, is heated to 40 °C. Determine thevolume of water at 40 °C if the coefficient of volumetric expansion ofwater within this range is 30 x 10-9 °C.
Given data
*The given initial volume of water is V_1 = 2 L
*The given initial temperature of the water is T_i = 20 °C
*The given final temperature of the water is T_f = 40 °C
*The coefficient of volumetric expansion of water is
[tex]\alpha=30\times10^{-9}^{}\text{C}[/tex]The formula for the volumetric expansion is given as
[tex]\frac{\Delta V}{V}=\alpha(T_f-T_i)[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \frac{\Delta V}{V_1}=30\times10^{-9}\times(40^0-20^0) \\ V_2-V_1=6\times10^{-7}V_1 \\ V_2=6\times10^{-7}V_1+V_1^{} \\ =V_1(6\times10^{-7}+1) \\ =2.00\text{ L} \end{gathered}[/tex]The element lanthanum has an isotope with an atomic mass of 138. 9 amu and an abundance of 99. 91%. A second isotope has an atomic mass of 137. 9 amu and an abundance of 0. 9%. What is the average atomic mass of lanthanum?.
The average atomic mass of lanthanum that has an isotope with an atomic mass of 138.9 amu and an abundance of 99.9 1% and a second isotope with an atomic mass of 137.9 amu and an abundance of 0.09 % is
Average atomic m = Sum of product of atomic m and % abundance / 100
For isotope 1,
Atomic mass = 138.9 amu
% abundance = 99.91 %
For isotope 2,
Atomic mass = 137.9 amu
% abundance = 0.09 %
Average atomic mass = ( 138.9 * 99.91 ) + ( 137.9 * 0.09 ) / 100
Average atomic mass = 13877.5 + 12.41 / 100
Average atomic mass = 138.9 amu
The average atomic mass or average atomic weight of an element is the weighted average mass or weight of atoms in a naturally occurring sample ( i.e., isotopes ) of the element.
Therefore, the average atomic mass of lanthanum is 138.9 amu
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first, launch the video solenoids. after watching the video, answer the follow-up question below. select to launch video part a a solenoid with 200 loops is 55 cm long. it has a current of 1.2 a . what is the magnetic field in this solenoid?
The magnetic field in the solenoid due to current 1.2A is 54.807 x 10⁻⁵ A/m
The No. of Loops in the solenoid, N = 200
Length of the solenoid, l = 55 cm
The current passing through the solenoid, I = 1.2 A
The magnetic field passing through the solenoid can be calculated using the Ampere's law which is
B = μ₀nI
where B is the magnetic field
μ₀ is the permeability of free space
n is the no. of turns per meter
I is the current passing through the solenoid
Thus , we can re-write this formula a
[tex]B = \frac{\mu_0 N I}{l}[/tex]
Let us substitute the know values in the above equation , we get
B = (4 x 3.14 x 10⁻⁷ x 200 x 1.2 ) / (55 x 10⁻²)
where μ₀ = 4π x 10⁻⁷
B = 3014.4 x 10⁻⁷ / 55 x 10⁻²
B = 54.807 x 10⁻⁷ x 10²
B = 54.807 x 10⁻⁵ A/m
Therefore , the magnetic field passing through the given solenoid is
54.807 x 10⁻⁵ A/m
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