Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.

Compute the PE when the spring is compressed by 0.50 m.

Answers

Answer 1

Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m


Related Questions

A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?

Answers

Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

Answers

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J

A car starting at rest accelerates at 3m/seconds square How far has the car travelled after 4s?​

Answers

Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?

Answers

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated in the patellar tendon is 400 N, what is the resulting angular acceleration, in rad/s2), if the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm

Answers

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

A Man has 5o kg mass man in the earth and find his weight​

Answers

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?

a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features

Answers

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?

Answers

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

the product 17.10 ✕

Answers

Explanation:

pls write the full question

An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________

a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.

Answers

Answer:

b. approaches zero.

Explanation:

The phenomenon is known as length contraction.

Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.

let the length of the train = L

Let the length observed when the train is in motion = L₀

Apply Einstein's special theory of relativity;

[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]

from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)

As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L).  (L₀ < L)

Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero.  (L₀ = 0)

Thus, the length of this object, as measured by a stationary observer approaches zero

The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off

Answers

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m[tex]_g[/tex]g = 0

where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m[tex]_g[/tex]g = 0

solve for m[tex]_g[/tex]

m[tex]_g[/tex] = ( Vρg - Mg ) / g

m[tex]_g[/tex] =  Vρg/g - Mg/g

m[tex]_g[/tex] =  ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m[tex]_g[/tex] / μ

where μ is the molecular mass of air

so

PV = (m[tex]_g[/tex]/μ)RT

solve for T

μPV = m[tex]_g[/tex]RT

T = μPV / m[tex]_g[/tex]R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T =  1405920 / 1979.442

T =  710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

The temperature required for the air to be heated is 710.26 K.

Given data:

The mass of a hot air-balloon is, m = 381 kg.

The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].

The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].

The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].

The condition where the hot air balloon is just about to take off is as follows:

[tex]F-mg - m'g =0[/tex]

Here,

m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,

[tex]F = V \times \rho \times g[/tex]

Solving as,

[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]

Now, apply the ideal gas law as,

PV = nRT

here, R is the universal gas constant and n is the number of moles and its value is,

[tex]n=\dfrac{m'}{M}[/tex]

M is the molecular mass of gas. Solving as,

[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]

Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,

[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]

Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.

Learn more about the ideal gas equation here:

https://brainly.com/question/18518493

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.0 m/s and the ball is 0.310 m from the elbow joint, what is the angular velocity (in rad/s) of the forearm

Answers

Answer:

[tex]\omega=109.67\ rad/s[/tex]

Explanation:

Given that,

The speed of the ball, u = 34 m/s

The ball is 0.310 m from the elbow joint.

We need to find the angular velocity (in rad/s) of the forearm.

We know that,

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]

So, the required angular velocity of the forearm is 109.67 rad/s.

Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way

Answers

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN

Answers

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be

Answers

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Do all substances conduct heat ?Why/ Why not ?​

Answers

Answer:

no, all substances doesnot conduct heat

Answer:

No, all substances do not conduct heat easily because it depends on the nature of the substance. Some are good conductors of heat and some are bad. Therefore, it depends on their characteristics and their ability to conduct heat.

The bad conductors of heat are water, air, plastic, wood, etc.

Gold, Silver, Copper, Aluminium, Iron, etc. are good heat conductors as well as electrical conductors.

An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x

Answers

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD

Answers

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank

Answers

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

           

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s

Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.

Required:
Rank from smallest to largest.

Answers

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic?

Answers

Answer:

The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic? is your ansewer dont take tension

The string will be 108 cm long when it is stretched to the point where it becomes plastic.

What is elasticity?

Elasticity in physics and materials science refers to a body's capacity to withstand a force that causes distortion and to recover its original dimensions once the force has been withdrawn.

When sufficient loads are applied, solid objects will deform; if the material is elastic, the object will return to its original size and shape after the weights have been removed. Unlike plasticity, which prevents this from happening and causes the item to stay deformed,

Given parameters:

The elastic extensibility of a piece of string is 0.08.

The string is 100 cm long.

Hence,  it becomes plastic, after  it is stretched up to = 100 × 0.08 cm = 8 cm.  The string will be 108 cm long.

Learn more about elasticity here:

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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.

Answers

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?

Answers

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.

Answers

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022


Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.

Answers

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?​

Answers

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n14.0 g/mol

describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please

Answers

Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.

Answers

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

Question 1 of 10
Which nucleus completes the following equation?
239UHe+?
A. 228 Th
B. 2220
c. 23. Pu
D. 78Th
SUBMIT

Answers

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time

Answers

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

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