A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?

Answers

Answer 1

A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s

From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.

i.e.

[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]

Given that:

the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m

From the equation above, multiplying both sides with 2, we have:

[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]

Making (k) the subject of the formula;

[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]

[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]

k = 216.26 N/m

However, when compressed to one-half of the maximum distance; the speed is computed as follows:

x = 0.68/2 = 0.34 m

[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]

[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]

[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]

25 - v² = 6.25

25 -6.25 = v²

v² = 18.75

[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]

v = 4.33 m/s

Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s

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Related Questions

_____is the nitrogen base found only in DNA.

Answers

Answer:

thymine

Explanation:

The nitrogen base found only in DNA is known as thymine which is also called 5-methyl uracil as thymine is a derivative of uracil.

A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration

Answers

Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .

Which of the
following
DECREASES
as you go UP a
mountain?
A. climate
B. altitude
C. amount of oxygen
D. buoyancy

Answers

C. Amount of oxygen

The others either change but don’t decrease or they increase.

Answer:

C. Amount of oxygen

Explanation:

Options A and D are invalid as they aren't affecting factors.

Option B is false as the altitude increases as you go up a mountain.

Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.

What is Newton's scientific view?

Answers

Answer:

Newton's first law of motion concerns any object that has no force applied to it.

Explanation:

three laws of motion and the law of universal gravitation.

If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron

Answers

Answer:

[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field

Explanation:

B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]

[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]

v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]

Magnetic force is given by

[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]

The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.

I’ll mark you as brinlist please help.

Answers

Answer:

245 divided by 5.14=47.6653696 or 47.66

Explanation:

47.6653 is the correct answer

which particle have a mass of 1 u​

Answers

Answer:

Explanation:

proton

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm

Answers

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t[tex]_{min[/tex] = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t[tex]_{min[/tex] = 750 / 2(1.20)

t[tex]_{min[/tex] = 750 / 2.4

t[tex]_{min[/tex] = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t[tex]_{min[/tex] = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t[tex]_{min[/tex] = 750 / 4(1.50)

t[tex]_{min[/tex] = 750 / 6

t[tex]_{min[/tex] = 125 nm

Therefore, the minimum thickness be now will be 125 nm


An object is projected from a height of 100m above the ground at an angle of 300to the
horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)​

Answers

Answer:

a) y = 127.6 m, b) 11.9s,  c)  v = 103.6  m / s, θ’= 326.7º

Explanation:

This is a missile throwing exercise

let's start by breaking down the initial velocity

         sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]

         cos 30 = v₀ₓ / v₀

         v_{oy} = v₀ go sin 30

         v₀ₓ = v₀ cos 30

         v_{oy} = 100 sin 30 = 50 m / s

         v₀ₓ = 100 cos 30 = 86.6 m / s

a) the maximum height is requested.

At this point the vertical velocity is zero (v_y = 0)

         v_y² = [tex]v_{oy}^2[/tex] - 2 g y

         0 = v_{oy}^2 - 2g y

         y = [tex]\frac{v_{oy}^2 }{2g}[/tex]

         y = 50² / (2  9.8)

         y = 127.6 m

b) Flight time

this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m

         y = y₀ + v₀ t - ½ g t²

         

       0 = 100 + 50 t - ½ 9.8 t²

we solve the quadratic equation

        4.9 t² - 50 t - 100 = 0

         t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]

         t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]

         t₁ = 11.9 s

         t₂ = -8.4 s

flight time is 11.9s

c) The time 1 s before hitting the ground is

        t1 = 11.9 -1

        t1 = 10.9 s

let's find the vertical speed

       v_y =[tex]v_{oy}[/tex] - g t

       v_y = 50 - 9.8 10.9

       v_y = -56.8 m / s

the negative sign indicates that the direction of the velocity is downward.

On the x-axis there is no acceleration therefore the speed is constant.

Let's use the Pythagorean theorem

        v = [tex]\sqrt{v_x^2+v_y^2}[/tex]

        v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]

        v = 103.6  m / s

let's use trigonometry

        tan θ = [tex]\frac{v_y}{v_x}[/tex]

        θ = tan⁻¹  \frac{v_y}{v_x}

        θ = tan⁻¹  (-56.8 / 86.60)

        θ = -33.3º

the negative sign indicates that it is measured clockwise from the x-axis

for a counterclockwise measurement

          θ’= 360 - θ

          θ' = 360 - 33.3

          θ’= 326.7º


Please help me easy question just get mixed up.

Answers

Answer:

The mechanical energy of the ball is 112.667 J

Explanation:

Mechanical energy = kinetic energy + potential energy

Where:

kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]

potential energy = mgh

So that for the given ball, its mechanical energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] + mgh

                                = m([tex]\frac{1}{2}[/tex][tex]v^{2}[/tex] + gh)

But, m = 400 g = 0.4 kg, v = 23.0 m/s, g = 9.81 m/[tex]s^{2}[/tex] and h = 1.75 m.

Mechanical energy of the ball = 0.4([tex]\frac{1}{2}[/tex]([tex]23^{2}[/tex]) + 1.75*9.81)

                                                  = 0.4(264.5 + 17.1675)

                                                  = 0.4 x 281.6675

                                                  = 112.667

The mechanical energy of the ball is 112.667 J.

Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region

Answers

Answer:

I'd say D

Explanation:

because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)

Pls help will mark Brainliest

Answers

I want to say the answer is 100

Select the correct answer Which object is an insulator

A. iron
b. cooper
c. plastic
d. salt water​

Answers

c. plastic. plastic is an insulator!
C is the answer to the question

Blocks A (mass 5.00 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 3.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line.
(a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
(b) Find the velocity of each block after they have moved apart.

Answers

Answer: i believe  its B Find the velocity of each block after they have moved apart sorry

Explanation: have a nice day buddy

Pls help ASAP
Imagine that Maritans launch a rocket toward the Earth at a great speed. While the
rocket is traveling toward us, it will appear
than it actually is.
O more blue
darker
larger
more red

Answers

Answer:

The rocket will appear larger than it actually is

What is the importance of using locally available resources in creating art?​

Answers

Answer:

please give me brainlist and follow

Explanation:

Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.

HELP URGENT PLEASE!!!!!!!

Answers

Answer:

I think c I dont know sorry if I'm wrong

It is C because the North Magnets are stronger than the south magnets and C is the only diagram showing that relation in the middle :)

Which of these cubes absorb the most light?

Answers

Answera black cube or dark colors cause dark colors suck in heat

A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.80 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 87.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j direction is upward.)

Answers

The normal force exerted by the floor of the elevator on the student during her brief acceleration is 1357.2 N.

To determine the normal force exerted by the floor of the elevator on the student during the brief acceleration, we need to analyze the forces acting on the student.

The force of gravity acting on the student can be calculated using the equation F_gravity = m * g, where m is the mass of the student and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, F_gravity = 87.0 kg * 9.8 m/s² = 852.6 N.

Since the elevator and the student are accelerating upward, there is an additional upward force acting on the student. This force is equal to the mass of the student multiplied by the acceleration of the elevator, which is given as 5.80 m/s². So, the upward force is F_upward = m * a = 87.0 kg * 5.80 m/s² = 504.6 N.

Now, the normal force exerted by the floor of the elevator on the student is equal to the sum of the gravitational force and the upward force. In this case, the normal force is given by F_normal = F_gravity + F_upward = 852.6 N + 504.6 N = 1357.2 N.

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7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.​

Answers

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque

Answers

Your answer is c, work

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

Answers

Answer:

  x = 0.455 L

Explanation:

For this exercise we must use the rotational equilibrium condition

        Σ τ = 0

it has two forces, the first is perpendicular to the rod, so its stub is

         τ₁ = F₁ L

the second force is applied with an angle, so we can use trigonometry to find its components

          sin θ = F_parallel / F₂

          cos θ = F_perpendicular / F₂

         F_parallel = F₂ sin θ

         F _perpendicular = F₂ cos θ

torque is

         τ₂ = F_perpendicular x + F_parallel 0

the parallel force is on the rod therefore its distance is zero

           

we apply the equilibrium equation

          τ₁  - τ₂ = 0

          F₁ L = F₂ cos θ  x

          x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]

let's calculate

          x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]

          x = 0.455 L

light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and second (m=2) interference minima?

Answers

Answer:

3.35

Explanation:

Got it on Acellus

The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.

What is diffraction?

Diffraction is the phenomenon of bending of waves through obstacles.  

Given is the wavelength λ= 485 nm, silt width d =  8.32 *10⁻⁶ m, then the angle θ will be

d sinθ =mλ

for m=1, sin θ₁ = λ/d

for m=2, sin θ₂ = 2λ/d

Substitute the values into both expressions to find the angles,

sin θ₁ = 485 x 10⁻⁹ /  8.32 *10⁻⁶

θ₁ = 3.34°

and sin θ₂ = (2 x  485 x 10⁻⁹ )/  8.32 *10⁻⁶

θ₂ = 6.7°

The angle between m =1 and m=2 will be

θ₂ -θ₁ = 6.7° - 3.34° =3.36°

Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.

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What do thermal energy and electrical energy have in common

Answers

Answer:

you can write some points its an explanation

and similarities. or common

Explanation:

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy. A whole branch of physics, thermodynamics, deals with how heat is transferred between different systems and how work is done in the process (see the 1ˢᵗ law of thermodynamics).

The faster the atoms or molecules move, the more heat or thermal energy they have. ... A hair straightener turns the electrical energy from a wall outlet into heat (thermal energy). 4. As electricity runs through the filaments in a space heater, the electrical energy is converted into heat (thermal energy).

An object undergoing simple harmonic motion takes 0.15 s to travel from one point of zero velocity to the next such point. The distance between those points is 30 cm. (a) Calculate the period of the motion. s (b) Calculate the frequency of the motion. Hz (c) Calculate the amplitude of the motion. cm

Answers

Answer:

Explanation:

Point of zero velocity are extreme points situated on either side of equilibrium position .

a )

Time taken to travel between these two points is .15 s

time for half the oscillation = .15 s

Time for full one oscillation = .30 s .

Time period of oscillation = .30 s

b)

frequency = 1 / time period

= 1 / .30s = 3.33 oscillation per second.

c )

Distance between these two point is equal to two times amplitude

2 x amplitude = 30 cm

amplitude = 15 cm

A water balloon weighing 4.5 N rests on a table. The balloon has an area of 2.6 x 10-3
m² in contact with the table. What pressure does the balloon exert on the table?

Answers

Answer:

the pressure the balloon exerts on the table is 1,730.77 N/m²

Explanation:

Given;

weight of the water balloon, F = 4.5 N

area of the balloon, A = 2.6 x 10⁻³ m²

The pressure the balloon exerts on the table is calculated as follows;

[tex]P = \frac{F}{A}[/tex]

substitute the given values and solve for pressure, P;

[tex]P = \frac{4.5}{2.6 \times 10^{-3}} \\\\P = 1,730.77 \ N/m^2[/tex]

Therefore, the pressure the balloon exerts on the table is 1,730.77 N/m²

Please answer this for 15 points please don’t put in a link.

Answers

It’s b I took the test

Answer:

c. Double Replacement

Explanation:

As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.

Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2

and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.

Review please help.

Answers

Answer:

1 and 3

Explanation:

because they are going up from 0

Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.

Answers

Explanation:

As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.

if the density of a napthalene ball is 0.02kg.what is the mass of the napthalene ball if it has a volume of 100m³

Answers

Density=mass/volume
Mass=100x0.02=2
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