The block will oscillate back and forth along the horizontal surface with a period given by the equation $T = 2\pi\sqrt{\frac{m}{k}}$,
where $m$ is the mass of the block, $k$ is the force constant of the spring, and $T$ is the period of the oscillation. In this case,
the mass of the block is 150 g, which is equivalent to 0.15 kg, and the force constant of the spring is 3.2 N/m.
Plugging these values into the equation, we get
$$T = 2\pi\sqrt{\frac{0.15}{3.2}} \approx 0.3268\text{ s}$$
The amplitude of the oscillation is equal to the maximum displacement of the block from its equilibrium position, which in this case is 3.0 cm. The block will oscillate back and forth with this amplitude and with a period of approximately 0.3268 seconds.
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hree children are riding on the edge of a merry-go-round that is 115 kg, has a 1.7-m radius, and is spinning at 24 rpm. the children have masses of 32, 34, and 40 kg. (a) if the children with masses of 34 and 40 kg move to the center of the merry-go-round, what is the new angular velocity in rpm?
The new angular velocity in rpm is
27.7 rpm
M = 115 kg be the mass of the merry-go-round,
r = 1.7 m be its radius,
m1 = 32 kg,
m2 = 34 kg,
m3 = 40 kg be the masses of the three children,
ω = 22 rpm be the angular velocity with all children at the edge,
ω' (to be calculated) be the angular velocity after m2 is moved to the center.
The conservation of angular momentum can be used to resolve this.
The angular momentum before and after the move must be equal.
But there are two presumptions we must make.
1) The merry-go-round has a consistent disc shape. Although it is improbable, if the merry-go-round has a more widespread. The issue gets more complex due of its convoluted shape.
2) The kids are on the merry-go-outer round's edge. Regardless of where the kids are, the issue is easily fixable.
but, since I have no information, I'll guess it's on the edge.
In general, a mass m at the edge contributes mr²ω to the angular momentum.
The rotating uniform disk contributes Mr²ω/2 to the angular momentum.
Now we can equate the total angular momentum after the move to the total
angular momentum before the move.
Mr²ω'/2 + m1r²ω' + m3r²ω' = Mr²ω/2 + m1r²ω + m2r²ω + m3r²ω
From that express
ω' = ω(M/2 + m1 + m2 + m3)/(M/2 + m1 + m3)
Substituting actual numbers
ω' = 22×(115/2 + 32 + 34 + 40)/(115/2 + 32 + 40) = 27.7 rpm
Hence, new angular velocity in rpm is 27.7 rpm
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6. conical surface of constant density find the moment of iner?tia about the z-axis of a thin shell of constant densthe cone 4x 2 4y 2 - z 2
We need to find the rotational inertia of the shell about the axis of rotation.
What is rotational inertia?Inertia of rest the inability of a body to change by itself its state of rest is called inertia of rest. Inertia of direction. The inability of a body to change by itself its direction of motion. Inertia of motion: The inability of the body to change by itself its state of motion is called inertia of motionYou tend to move forward when a sudden break is applied.You feel a backward force when the bus moves quickly from rest.Dusting bed with a broom removes dust due to inertia of rest.when you shake a branch the leaves get detached.According to Newton's first law of motion, an object with a given velocity maintains that velocity unless acted on by an external force. Inertia is the property of matter that makes this law hold true.Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be.When a car makes a sharp turn at a high speed, the driver tends to get thrown to another side due to inertia of direction.The spark coming out of a grinding stone is tangential to the rotating stone due to directional inertia.The mud from the wheels of a moving vehicle flies off tangentially.[tex]I_{y} = (\frac{1}{8}) \pi r^{4}[/tex]
x = r cos θ
y = r sin θ
dA = r dr dθ
Then the moment of inertia about the x-axis will be
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The abundance of __________________ _________________( H, He, Li) in the universe is almost exactly as what The Big Bang predicts. Elements were first fused within the _______________________ after the Big Bang
The abundance of light elements such as Hydrogen (H), Helium (He), and Lithium (Li) in the universe is almost exactly as what the Big Bang predicts. Elements were first fused within the protons and neutrons after the Big Bang.
What is Big bang theory?Light elements like hydrogen (H), helium (He), and lithium (Li) are abundant throughout the cosmos, almost exactly as predicted by the Big Bang theory. After the Big Bang, protons and neutrons were the first to combine elements. The most popular theory for how the universe came into being is called the Big Bang Theory. Simply put, it asserts that the universe as we currently understand it originated from a single point that was infinitely hot and dense and that it first expanded and contracted at unfathomable speeds before slowing down and expanding over the course of 13.7 billion years to become the universe that we see today.
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suppose that 20 j of work is needed to stretch a spring from a0 cm to 2a0 cm. 80 j of work is needed to stretch it from a1 cm to 2a1 cm. what is the natural length of the spring. note that a0 and a1 are some constants and the unit is cm instead of m
The needed length of spring or the natural length is ( 12a₀²- a₁²)( 8a₀- 2a₁).
Let L be the natural length,
Since 20J of work is demanded to stretch the spring from a₀ cm to 2a₀ cm,
We have,20 = k / 2(( 2a₀- L) ²-( a₀- L) ²)--------- eq 1
Again,80 J of work is demanded to stretch the spring from a₁ cm to 2a₁ cm,
80 = k / 2(( 2a₁- L) ²-( a₁- L) ²)---------- eq 2
Now, eq 2 / eq 1
k/ 2(( 2a₀- L) ²-( a₀- L) ²) k/ 2(( 2a₁- L) ²-( a₁- L) ²)
= (80/20)a₁( 3a₁- 2L)/ a₀( 3a₀- 2L)
= 43a₁ ²- 2La₁/ 3a₀ ²- 2La₀ = 43a₁ ²- 2La₁
= 12a₀ ²- 8La₀L( 8a₀- 2a₁)
= 12a₀ ²- 3a₁ ²L
= ( 12a₀ ²- 3a₁ ²)( 8a₀- 2a₁)
Therefore the needed length of spring or the natural length is ( 12a₀²- a₁²)( 8a₀- 2a₁).
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A ball is initially moving at 12 m/s up a ramp. The acceleration of the ball is -2 m/s/s down the ramp. What is the balls velocity after 8 seconds? (frame of reference + is up the ramp)
The ball is moving with a velocity of 28 m/s after 8 seconds.
What is the equation of motion?The equations of motion can be used to set up the relationship between the initial velocity and final velocity, acceleration, time, and displacement of a moving object.
The three equations of motion for a moving object can be represented:
[tex]v = u +at\\S = ut +\frac{1}{2} at^2\\v^2 - u^2 = 2aS[/tex]
Give, the initial velocity of the ball moving up, u = 12 m/s
The acceleration of the ball up the ramp, a = + 2m/s²
The final velocity can be calculated from the first equation of motion:
v = u + at
v = 12 + (2) (8)
v = 12 + 16
v = 28 m/s
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a 0.160-g puck strikes and embeds itself in a 6.4 kg block. the block moves off at 1.2 m/s. what was the puck's speed?
The puck's speed was 7.2 m/s.
What is speed?
Despite their similarities, speed and velocity have very different meanings, just as distance as well as displacement do. Speed, which is a scalar quantity, is the "speed at which an object is moving." The rate that an object travels a distance can be assumed of as its speed. A fast-moving object travels at a high speed and completes a significant distance in a brief period of time. This is in contrast to an object moving slowly, which travels a relatively short distance within the same amount of time. Zero speed refers to an object that is not moving at all.
The puck's speed can be determined by solving the equation for momentum conservation. Momentum is conserved when the total momentum before an interaction is equal to the total momentum after the interaction. In this case, the equation looks like this:
m_P * v_P + m_B * v_B = (m_P + m_B) * v_F
Where m_P is the mass of the puck, v_P is the initial velocity of the puck, m_B is the mass of the block, v_B is the initial velocity of the block, and v_F is the final velocity of the block.
Plugging in the given values, we get:
0.160 * v_P + 6.4 * 1.2 = (0.160 + 6.4) * 1.2
Solving for v_P, we get:
v_P = 7.2 m/s
Therefore, the puck's speed was 7.2 m/s.
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a circular coil of wire has a radius of 7 cm and contains 40 turns. a field of 0.900 t passing through the coil collapses at a uniform rate to zero in 0.200 seconds. what emf is generated in the wire?
The EMF generated by the coil is 2.772 V.
Radius of the coil = R = 7 cm = 0.07 m
Number of turns = N = 40
Initial Magnetic Field = B₁ = 0.900 T
Final Magnetic Field = B₂ = 0
Time = 0.2 s
Change in the Magnetic field of the coil =
= dB/dt = (0.9 - 0) / 0.2
= dB/dt = 9/2 T/s
Area of the coil = A = πR²
The total flux related to the circular coil = ∅
= ∅ = NBA
EMF induced by the coil =
= E = d∅/dt
= E = NA dB/dt
= E =40 X π X (0.07)² X (9/2)
= E =2.772 V
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within the deep cold zone, average temperature and salinity . a. remain fairly constant with depth. b. increases and decreases, respectively. c. decrease with depth. d. increase with depth.
The average salinity in the deep cold zone increases with depth. The average temperature in the deep cold zone decreases with depth.
Salinity in the ocean is how much salt is in ocean water. Near the surface, the salinity could change.
Salinity will decrease after rainfall.Salinity will increase after evaporation on a sunny day.Sunlight can affect the temperature of the water. But only about 200 m from the surface. Below 200 m, the sunlight cannot get through.
According to hydrostatic pressure, as deeper as the ocean, the pressure increase. When the pressure increase, the temperature will decrease.
In the deep cold zone, there is a water current called thermohaline circulation which carries colder and saltier water to the deeper parts of the ocean. This circulation makes in the deep cold zone the temperature decrease and the salinity increase.
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328-kg car moving at 19.1 m/s in the x direction hits from behind a second car moving at in the same direction. if the second car has a mass of and a speed of right after the collision, what is the velocity of the first car after this sudden collision?
The velocity of the first car after this sudden collision is +14 m/s.
What is velocity?
Velocity can be defined as the rate of change of position of object with respect to time or a specific frame of reference. it can also be defined as the speed of an object in a specific direction, this means that the value of velocity changes as the direction of the the movement of the object changes. this makes velocity a vector quantity.
Its SI unit is [tex]m/s^{-1}[/tex].
What are vector quantities?
Vector quantities are those quantities that have both the direction and the magnitude, eg: velocity, displacement, force etc.
According to the given information:
we know, mV + Mv = mU + Mu
where m = 320kg, M = 720 kg, v = 13m/s, V = 19 m/s, u = 15.1 m/s
U = (mV + Mv - Mu)/m
= (328 x 19.1 + 790 x 13 - 790 x 15.1)/328
= +14 m/s
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I understood the question to be:
A 328-kg car moving at 19.1 m/s in the + x direction hits from behind a second car moving at 13.0 m/s in the same direction. If the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision?
the maximum speed of a child on a swing is 4.9 m/s. the child's height above the ground is 0.70 m at the lowest point in his motion. how high above the ground is he at his highest point
1.925 m high above the ground is he at his highest point.
As per the question,
V=4.9 m/s
h= 0.70 m (child's height above the ground at the lowest point in the motion)
Height of the child above the ground at his highest point=h’
H=h’-h
1/2mV²=mgH (g=9.8m/s²)
V²=2gH [V=2g(h’-h)]
V=√ 2×9.8×(h’-0.70)
(4.9)²=19.6(h’-0.70)
24.01=19.6h’-13.72
37.73=19.6h’
37.73/19.6=h’
1.925=h’
What is motion?In physics, motion is defined as a change in a body's position or orientation over time. Translation is the term for motion along a line or a curve. Rotation is another motion that modifies a body's orientation.
What is periodic motion?Periodic motion is defined as a motion that repeats at regular periods of time.
The motion of a youngster on a swing is periodic because the motion of a swing also repeats after short intervals.
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What is the formula for the compound that forms between magnesium and phosphorous?
Answer:
Magnesium phosphate | Mg3(PO4)2
Explanation:
The formula for the compound that forms between magnesium and phosphorous is Mg3(PO4)2
a small artery has a length of 1.6 ✕ 10−3 m and a radius of 2.4 ✕ 10−5 m. if the pressure drop across the artery is 1.8 kpa, what is the flow rate (in mm3/s) through the artery? (assume that the temperature is 37°c.)
The flow rate (in mm3/s) through the artery is [tex]7.031 * 10^-^1^1 m^3/s[/tex]
What is Poiseuille's Law?
Poiseuille's Law can be used to explain the fluid flow through an IV catheter. According to this, the viscosity (μ), pressure gradient (P), length (L), and diameter (r) of the tubing are all important variables that affect fluid flow (Q).
It is given by equation
Δp = 8μLQ / π[tex]R^4[/tex]
where,
Δp = pressure difference between the two ends
μ = dynamic viscosity
L = length of pipe
Q = volumetric flow rate
π = pi
R = pipe radius
Given,
Length of artery L = [tex]1.6 * 10^-^3m[/tex]
Radius R = [tex]2.4 * 10^-^5m[/tex]
Pressure Δp = 1.8 kPa = [tex]1.8 * 10^3 Pa[/tex]
Viscosity μ = [tex]2.084 * 10^-^3 Pa. s[/tex]
Subsituting the values in equation,
Δp = 8μLQ / π[tex]R^4[/tex]
Q = Δp × π[tex]R^4[/tex] / 8μL
Q = [tex]\frac{(1.8 * 10^3) (3.14)(2.4 * 10^-^5)^4}{8 * (2.084* 10^-^3)(1.6 * 10^-^3)}[/tex]
Q = [tex]7.031 * 10^-^1^1 m^3/s[/tex]
The flow rate (in mm3/s) through the artery is [tex]7.031 * 10^-^1^1 m^3/s[/tex]
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A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 8 m from the takeoff point. part a if the kangaroo leaves the ground at a 19 ∘ angle, what is its takeoff speed
The take-off speed of the gray kangaroo is 11.28 m/s.
How to calculate the takeoff speed?The kangaroo's final speed is zero when it is at its highest point above the ground, and the time it spends there is halved from the time it will take it to hit the earth again.
To calculate the horizontal distance (S) travelled by the kangaroo we need the equation,
[tex]S=v_{0} cos\theta t[/tex]
Here,
v₀ = takeoff speed
[tex]\theta[/tex] = the angle of projection
t = total time taken by the kangaroo.
Given,
The take-off distance for each jump attained by the kangaroo = 8 m
The angle of projection = 19°
Substituting the above values in the above equation,
[tex]8 = v_{0} cos19t[/tex]
[tex]t=\frac{8}{v_{0}cos19 }[/tex]
[tex]t=\frac{8}{0.9455v_{0} }[/tex]
The following equation can be used to determine the kangaroo's final speed when it reaches its highest point above the ground:
[tex]v=v_{0} sin\theta-\frac{gt}{2}[/tex]
Now, Substitute 0 for the value of v we get,
[tex]0=v_{0} sin\theta \frac{g}{2} ( \frac{8}{0.9455v_{0} })\\v_{0} = \sqrt{\frac{8g}{2*0.9455*sin\theta} } \\[/tex]
[tex]v_{0} = \sqrt{\frac{8g}{2*0.9455*sin\theta} } \\\\v_{0} = \sqrt{\frac{8*9.80}{2*0.9455*sin19} }\\v_{0}= 11.28 m/s[/tex]
Hence, the take-off speed of the gray kangaroo is 11.28 m/s.
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what type of conduit, fittings, switches, lighting fixtures, and receptacles are to be used in the flammable storage area?
The combustible storage room must use cable, connectors, switches, lighting, and plugs of the explosion-proof E12 variety.
What purpose serves a switch?Numerous devices can be connected using switches. Switches prevent route between two devices connected to the same network from obstructing the use of other devices. Switches let you manage who will have access to different network areas. Switches let you keep an eye on utilization.
Explain what switches are.A switch is a device used in electrical engineering that has the ability to connect or detach a circuit's carrying line, halting the flow of electricity, or rerouting it to another conductor.
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exhibit 6-5 the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. refer to exhibit 6-5. what percentage of items will weigh between 6.4 and 8.9 ounces? a. .2881 b. .1145 c. .4617 d. .1736
The percentage of items will weigh between 6.4 and 8.9 ounces 0.4617.
In science and engineering, the burden of an object is the force performing on the item because of gravity. a few well-known textbooks outline weight as a vector quantity, the gravitational force performed on the object. Others outline weight as a scalar quantity, the importance of the gravitational pressure.
Retaining your weight inside the ordinary range is an important part of healthful growing old. As in other tiers of existence, elevated body mass index (BMI) in older adults can growth the chance of developing fitness issues. those encompass heart ailments, excessive blood strain, stroke, and diabetes.
Bad sleep, sedentary activities, and consuming too many processed or sugary meals are simply some of the habits that could increase your hazard of weight advantage. yet, some simple steps — together with mindful ingesting, workout.
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I need a detailed answer
The mass of the beam, given that the beam is balanced is 400 g
How do I determine the mass of the beam?First, we shall re-conctruct the diagram given to better understand what we are looking for. Please see attached photo.
In the attached photo,
M is the mass of the beamNote: The mass of the beam act at the centre of the beam
Now, we shall determine the mass of the beam as follow:
Clock wise moment = M × 10Anti-clock wise moment = 200 × 20Mass of beam (M) =?Anti-clock wise moment = Clock wise moment
200 × 20 = M × 10
4000 = M × 10
Divide both sides by 10
M = 4000 / 10
M = 400 g
Thus, from the above calculation, we can conclude that the mass of the beam is 400 g
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A 30-cm-diameter, 1.4 kg solid turntable rotates on a 1.4-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 12 seconds. Part A How much friction force does the brake pad apply to the shaft?
0.628N friction force is applied by the brake pad to the shaft when a 30-cm-diameter, 1.4 kg solid turntable rotates on a 1.4-cm-diameter, 450 g shaft at a constant 33 rpm and when you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 12 seconds.
To determine the friction force applied by the brake pad to the shaft, first we need to calculate the moment of inertia of the solid turn table. Let's consider the moment of inertia of turn table as a disk then -
I = MR²/2
where M= mass of the turntable=1.4 kg
R= radius of the turntable
= diameter of the turntable/2
R= 30cm/2 = 15cm or 0.15m
substituting the value -
I = 1.4kg*(0.15)²/2
I = 0.0157kg-m²
When we hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 12 seconds so the acceleration (α) will be-
α = (ωf-ωi)/t
The initial angular speed (ωi) of the turntable = 33rpm or 3.45rad/s
The final angular speed (ωf) of the turntable = 0 rad/s
so,
α = (0 rad/s - 3.45 rad/s)/12s
α = -0.28rad/s²
Now, the frictional force applied by the brake pad is-
Torque = I x α
from dynamics of rotational motion
r x f = I x α
f = (I x α)/r
where r = radius of the distance from the pivot point so
1.4cm/2 or 0.007m
f = (0.0157kg-m²) * (0.28 rad/s²)/ 0.007m
f = 0.628N
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g capacity, the ability of the material/structure to safely resist load without breaking or damage, depends on?
Ductility determines whether a material or structure can safely withstand a load without breaking or suffering damage.
Ductility: What is it?A material's ductility is a mechanical property that refers to how easily it can be drawn from. In the field of materials science, ductility refers to how much plastic deformation under tensile stress a material can withstand before breaking. Ductility is a top priority in both engineering and manufacturing.
It describes a material's suitability for specific manufacturing processes (like cold working) and its capacity to withstand mechanical overload. Among the metals that are frequently referred to as ductile are copper and gold. Not all metals, like cast iron, can fail brittlely; some can instead fail ductilely. Polymers can be thought of as ductile materials because they typically allow for plastic deformation.
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at what speed does the rock strike the water if it is thrown with the same initial speed but at an angle 24 degrees below the horizontal?
The offered statement states that the rock hits the water at various times and with equal velocity.
What is a speed set of questions?The number of kilometers reached to the time needed to travel that distance is known as speed. Since speed simply has a part and no magnitude, it is a proportionality constant.
Briefing:
If we ignore air resistance due to the energy conservation, then the velocity of both stones would have to be equal because energy is a finite resource. They will each have the same sum of kinetic and potential energies at the beginning, hence the potential and the kinetic energies must be identical at the end, this happens when they reach the ground. Potential energy will change into kinetic energy.
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determine the moment of inertia of a 8.20 kg sphere of radius 0.793 m when the axis of rotation is through its center
The moment of inertia of the sphere when the axis of rotation is through its center be 2.063 kg-m².
What is moment of inertia?The sum of the products generated by multiplying each particle's mass by the square of its distance from the axis results in the moment of inertia (I), which is always expressed with respect to that axis.
Given parameters:
Mass of the sphere: m = 8.20 kg.
Radius of the sphere: r = 0.793 m.
Hence, the moment of inertia of the sphere when the axis of rotation is through its center be = 2/5 mr²
= 2/5 × 8.20 × 0.793 × 0.793 kg-m².
= 2.063 kg-m².
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you have been hired to design a spring-launched roller coaster that will carry two passengers per car. the car goes up a 13-m-high hill, then descends 20 m to the track's lowest point. you've determined that the spring can be compressed a maximum of 2.5 m and that a loaded car will have a maximum mass of 420 kg. for safety reasons, the spring constant should be 14 % larger than the minimum needed for the car to just make it over the top.
a) what spring constant (k) should you specify ?
b) What is the maximum speed of a 350 kg car if the spring iscompressed the full amount.
Based on safety considerations, the spring constant to be provided is computed to be k = 1661.4N/m. The speed of a 370 kg car when the spring is fully compressed is computed to be v = 14.6 m/s.
Where F = applied force and x = spring displacement, the formula for the spring constant is provided. The absence of a positive sign indicates that the restorative force opposes the speed.
In common use and kinematics, an object's speed (often abbreviated as v) is defined as the size of the change in position per unit of time or over time for an object.
These calculations are used to determine the car's weight:
W = mg W = 420* 9.8 = 4116 N
F = k(spring constant)
K = f=F/x = 4116/2.5 = 1646.4N/m is the formula for the spring constant.
Given the need for safety, the spring constant to be used is determined as follows: k = 1646.4N/m + 15 = 1661.4N/m.
When the spring is fully compressed, the 370 kg automobile will move at a speed of 14.6 m/s because 1/2 mv^2 = 1/2kx^2
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write an expression for the gravitational potential energy of the falling mass in terms of the angular displacement
The potential energy of a falling mass is [tex]PE=-m g R \theta[/tex] and [tex]g=9.8 \mathrm{~m} / \mathrm{s}^2.[/tex]
As we know,
The expression that relates linear displacement and angular displacement is given below:
[tex]s=r \theta[/tex]
Now we will replace s with h and r with R in the above expression.
[tex]h=R\theta[/tex]
Here, the height is h and the radius is R.
The initial potential energy is as follows:
[tex]P E_{\mathrm{i}}=m g h[/tex]
Now, replace h with R[tex]\theta[/tex].
We will get it as: [tex]P E_{\mathrm{i}}=m g R \theta[/tex]
Now, substituting [tex]$0^{\circ}$[/tex] for [tex]\theta[/tex]
[tex]P E_{\mathrm{i}} & =m g R\left(0^{\circ}\right) \\=0 \mathrm{~J}[/tex]
The expression for the final potential energy will become:
[tex]P E_{\mathrm{f}}=m g R \theta[/tex]
The potential energy of the falling mass will become:
[tex]P E=P E_{\mathrm{i}}-P E_{\mathrm{f}}[/tex]
From Expressions (1) and (2), the following expression can be obtained:
[tex]P E =0 \mathrm{~J}-m g R \theta \\=-m g R \theta[/tex]
The value of g is as follows: g=9.8 [tex]\mathrm{~m} / \mathrm{s}^2[/tex]
The potential energy of a falling mass is [tex]PE=-m g R \theta[/tex] and [tex]g=9.8 \mathrm{~m} / \mathrm{s}^2.[/tex]
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at the instant that the mass passes through the position where the spring is unstretched, what can be said about its instantaneous acceleration?
The instantaneous acceleration is maximum.
What is instantaneous acceleration?The main definition of instantaneous acceleration is. The ratio of velocity change during a given time interval such that the time interval equals zero.The velocity of an object at a specific point in time or along its path is referred to as instantaneous velocity. Acceleration that occurs instantly. It is defined as the limiting value of the average acceleration when the time interval t becomes very small.The result is the instantaneous acceleration expressed mathematically as a(t)=ddtv, which is the derivative of the velocity function v(t) (t). Thus, instantaneous acceleration is the derivative of the velocity function, just as velocity is the derivative of the position function.
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a ball of mass 1.50 kg is hung from a spring which stretches a distance of 0.6875 m. the spring constant of the spring (a scalar quantity) is n/m
The spring constant of the spring (a scalar quantity) is 21.38 N/m.
What is spring constant ?The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the larger the spring constant.
How do you calculate the spring constant?x represents the spring's displacement from its equilibrium position, and k is the spring constant in N/m.
The constant of proportionality k is called Coulomb's constant.
According to given information spring's displacement(x)= 0.6875 m and mass(m)=1.50kg
F=-K × x
K=f/x=mg/x=1.50×(9.8)/0.6875=21.38N/m
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a mass m on a frictionless table is attached to a hanging mass m by a cord through a hole in the table. find the speed with which m must move for m to stay at rest.
The speed with which m must move from M to stay at rest is √Mgr/m m/sec .Centripetal Force is responsible for the speed of mass m.
We are given that final state of mass M is rest. It means its velocity is zero. If velocity is zero it means that force acts on the mass M is balanced. When forces are balanced ,body will remain at rest as component of force and gravity cancel each other.
So, if mass M is at rest we need to identify what force are acting on the mass M.
Since mass M is directly perpendicular to acceleration due to gravity, therefore gravity will act some force on it.
Since mass M is moving in circle, due to its movement mass M feels centripetal force.
Therefore, centripetal force due to mass m=Force due to gravity
=>m×(v²)/r=Mg
=>v² = (Mg×r)/m
=>v=√Mgr/m
Hence, speed of mass m is √Mgr/m,
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(Complete question) is:
a mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table. find the speed with which m must move for M to stay at rest.
What is the complete factorization of 8x2 − 8x + 2?
The complete factorization is 2 (2x-1)².
What is factorization?Factorization is a way of dividing mathematical algebraic statements into their components. The components will produce the original phrase if we multiply them once more.
Factorization is a way of dividing mathematical algebraic statements into their components. If we multiply the components again, we get the original equation.
Factorising is the opposite of extending brackets. Fully factorizing an expression implies putting it in brackets and removing the greatest common factors. Finding the largest common factor between each item in the phrase is the simplest method of factorization.
Given,
= 8x² - 8x +2
= 8x² - (4+4)x +2
= 8x² - 4x + 4x +2
= 4x (2x -1) -2 (2x-1)
= (4x-2) (2x-1)
= 2 (2x-1) (2x-1)
= 2 (2x-1)²
Thus, complete factorization is 2 (2x-1)².
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combining the two values calculated above tells us the total momentum before the collision. the total momentum of cart 1 and cart 2 before the collision is 0.01 . if you were to do the same calculations to determine the momenta of each cart after the collision and then combine them, you would find that the total momentum of cart 1 and cart 2 after the collision is -0.022 . therefore, the change in momentum (change = final -initial) pf the whole sistem from before the coallision is -0.03. this is one trial. Which of the following statements is most true?
A)There must be a calculation error, because momentum should be conserved in this collision.
B)This demonstrates that momentum is not conserved in this collision.
C)This demonstrates that the motion detector is not precise enough to conduct this
experiment.
E)More trials will not improve the sensitivity of the motion detectors, so there is no point in
collecting more data.
F)Averaging the change in momentum for multiple trials can tell us whether momentum is not
conserved here or the motion detector is imprecise.
Answer:
b
Explanation:
the light we see from the most distant known quasar was emitted around 10 gyrs ago. what was happening with our sun (and therefore earth) 10 gyrs ago? what was happening with the milky way galaxy at this time?
10 billion years ago, the sun was likely in its early stages of formation. At this time, the Milky Way was still in the process of collapsing and starting to form stars.
What is Milky Way?
The galaxy that contains our Solar System is known as the Milky Way, and its name refers to how it appears to us on Earth: a hazy band of light in the night sky made up of stars that are too close together to be seen individually. The Milky Way is a disk-shaped structure, but when viewed from Earth, it looks like a band. In 1610, Galileo Galilei used his telescope for the first time to separate the band of light in to the individual stars. Most astronomers believed that the Milky Way held all of the universe's stars until the early 1920s.
Gas and dust clouds were beginning to collect, and the first stars and proto-planetary disks were beginning to form. The Milky Way was also likely forming its spiral arms, as the young stars began to move around the center of the galaxy. As the stars aged, they began moving further away, creating the spiral arms that we see today.
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a disk of mass 7 kg and radius 0.6m hangs in the xy plane from a horizontal low friction axle. the axle is 0.06m from the center of the disk. what is the frequency of f of small - angle oscillations of the disk ?
The frequency of of small - angle oscillations of the disk is 1.99/sec. Frequency is converse of time period.
We know very well that,
T=2π√I/mgh
where T is the time period,
I is the moment of inertia of body
m is the mass of the body
g is the acceleration due to gravity
and h is the height of the point from center of body
Now, we know very well that moment of inertia of disk is given by=(1/2)mr² where m is the mass of the body and r is the radius of the body
So, putting the values of mass and radius of disk, we get
=>I=(1/2)×7×(0.6)²
=>I=3.5×0.36
=>I= 1.26kg-m²
Now, h is given as 0.06m,value of g is 9.8m/sec² and value of m is 7kg
So, mgh=7×9.8×0.06=4.116kgm²/sec²
Therefore, T=2π√1.26/4.116
=>T=2π√0.306
=>T=2π×0.55sec
=>T=6.28×0.55sec
=>T=3.454sec
=>Frequency(ω)=2π/T
=>Frequency(ω)=2π/3.454
=>ω=6.28/3.454
=>ω=1.99/sec
Hence, frequency is 1.99/sec.
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a planet with a radius of 6.00 x 107 m has a gravitational field of magnitude 30.0 m/s? at the surface. what is the escape speed from the planet?
The escape speed of the planet is 756.60 m/s
How to calculate escape speed from the planet ?The earth's surface escape speed is roughly 11.2 km/s. This means that an object must have a minimum beginning velocity of 11.2 km/s in order to escape the gravity of the earth and journey into limitless space.
The minimum speed required for a free, unpropelled object to be able to escape the gravitational pull of a primary body and travel an infinite distance from it is known as the escape velocity or escape speed in celestial mechanics. Typically, it is expressed as an ideal speed that disregards atmospheric friction.
Given ,
Gravitational magnitude = 30.00m/s
Escape speed from planet = √[2×30/6*107][
√[2×30.00] /6 ×107
= 756.60 m/s
Therefore Escape speed from planet is 42.4264
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