An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring's maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring's maximum compression.
??m

Answers

Answer 1

(a) As it gets compressed by a distance x, the spring does

W = - 1/2 (52.1 N/m) x ²

of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so

W = ∆K

- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x0.118 m

(b) Taking friction into account, the only difference is that more work is done on the object.

By Newton's second law, the net vertical force on the object is

F = n - mg = 0

where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives

n = mg = 2.45 N

and from this we get the magnitude of kinetic friction,

f = µn = 0.120 (2.45 N) = 0.294 N

Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:

W (friction) = - (0.294 N) x

W (spring) = - 1/2 (52.1 N/m) x ²

==>   W (total) = W (friction) + W (spring)

Solve for x :

- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x0.112 m

Answer 2

For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:

a) The spring's maximum compression when the track is frictionless is 0.118 m.

b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.

 

a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex]  (1)

Where:

[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg

[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s

k: is the spring constant = 52.1 N/m

x: is the distance of compression

After solving equation (1) for x, we have:

[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]

Hence, the spring's maximum compression is 0.118 m.

b) When the track is not frictionless, we can calculate the spring's compression by work definition:

[tex] W = \Delta E = E_{f} - E_{i} [/tex]

[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex]   (2)

Work is also equal to:

[tex] W = F*d = F*x [/tex]     (3)

Where:  

F: is the force

d: is the displacement = x (distance of spring's compression)  

The force acting on the object is given by the friction force:

[tex] F = -\mu N = -\mu m_{o}g [/tex]   (4)

Where:

N: is the normal force = m₀g

μ: is the coefficient of kinetic friction = 0.120

g: is the acceleration due to gravity = 9.81 m/s²

The minus sign is because the friction force is in the opposite direction of motion.

After entering equations (3) and (4) into (2), we have:

[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]

[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]

[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]        

Solving the above quadratic equation for x

[tex] x = 0.112 m [/tex]  

Therefore, the spring's compression is 0.112 m when the track is not frictionless.

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An Object Moving Along A Horizontal Track Collides With And Compresses A Light Spring (which Obeys Hooke's

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A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?

Answers

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

[tex] A = l*w [/tex]

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]                            

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I hope it helps you!                        

Answer:

the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m

the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

[tex]A_l[/tex]= (L+0.5)(W+0.5)

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To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

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Answer:

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Answers

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Answer:

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Answer:

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Answer:

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Answers

Explanation:

Given:

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[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]

[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]

b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as

[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]

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Answers

Answer:

c) charges exert forces on other charges.

Explanation:

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Answers

The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.

To find the angle of a resultant vector, the vector must be resolved into y-component and x-component.

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[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\where;\\\\\theta \ is \ the \ direction \ of \ the \ resultant \ vetcor\\\\R_y \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ y - direction\\\\R_x \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ x - direction[/tex]

Therefore, the angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.

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Answer:

300J

Explanation:

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Answer:

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Edmentum

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube

Answers

Answer:

[tex]v=0.9833\ c[/tex]

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

[tex]$\text{Density} = \frac{m}{lwh}$[/tex]

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 [tex]kg/m^3[/tex]

So,

[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]

[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]

l = 0.024 m

Then for relativistic length contraction,

[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]

[tex]$\frac{v^2}{c^2}= 0.967$[/tex]

[tex]$\frac{v}{c}=0.9833$[/tex]

[tex]v=0.9833\ c[/tex]

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

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Answers

La energía que almacena el resorte cuando se comprime y estira 12 cm es 9,4 J.  

La energía potencial elástica del resorte se puede calcular con la siguiente ecuación:

[tex] E_{p} = \frac{1}{2}kx^{2} [/tex]

En donde:

k: es la constante del resorte = 1300 N/m

x: es la distancia de compresión o de elongación = 12 cm = 0,12 m

Dado que la energía es proporcional al cuadrado de la distancia recorrida por el resorte (x), la energía almacenada por el resorte durante la compresión será la misma que la energía almacenada por la elongación.

Por lo tanto, la energía almacenada es:

[tex]E_{p} = \frac{1}{2}kx^{2} = \frac{1}{2}1300 N/m*(0,12 m)^{2} = 9,4 J[/tex]                                                            

Entonces, la energía del resorte cuando se comprime y cuando se estira es la misma, a saber 9,4 J.                

Para saber más sobre energía potencial visita este link: https://brainly.com/question/156316?referrer=searchResults

Espero que te sea de utilidad!

Answer:

Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.

Explanation:

La Energía Potencial Elástica almacenada por el resorte ([tex]U_{e}[/tex]), en joules, se calcula a partir de la Ley de Hooke, la definición de Trabajo y el Teorema del Trabajo y la Energía, cuya expresión se presenta abajo:

[tex]U_{e} = \frac{1}{2}\cdot k\cdot (x_{f}^{2}-x_{o}^{2})[/tex] (1)

Donde:

[tex]k[/tex] - Constante elástica del resorte, en newtons por metro.

[tex]x_{o}[/tex] - Posición inicial del resorte, en metros.

[tex]x_{f}[/tex] - Posición final del resorte, en metros.

Nótese que el resorte sin deformar tiene una posición de cero, la tensión tiene un valor positivo y la compresión, negativo.

Asumiendo que en ambos casos el resorte se encuentra inicialmente sin deformar, se reduce (1) a una forma de función par, es decir, una función que cumple con la propiedad de que [tex]f(x) = f(-x)[/tex], se encuentra que al comprimirse o estirarse en la misma medida almacena la misma cantidad de energía.

La cantidad de energía a almacenar es:

[tex]U_{e} = \frac{1}{2}\cdot \left(1300\,\frac{N}{m} \right)\cdot (0,12\,m)^{2}[/tex]

[tex]U_{e} = 9,360\,J[/tex]

Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.

As the speed of a particle approaches the speed of light, the momentum of the particle Group of answer choices approaches zero. decreases. approaches infinity. remains the same. increases.

Answers

Answer:

approaches infinity

Explanation:

There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.

The relativistic momentum can be written as:

[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

where

u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.

m = mass of the object

c = speed of light.

So, as u tends to c, we will have:

[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.

Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.

A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?

Answers

Answer:

Solution

Verified by Toppr

Correct option is

C

3 cm

RI=apparent depthreal depth

Substituting, 34=apparentdepth12

Therefore, apparent depth=412×3=9

The height by which it appears to be raised is 12−9=3cm

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Answers

The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.

The factors that affect the intensity of radiation are

Type of emission from the source :This  can be alpha, gamma, beta or electromagnetic rays etc

Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.

Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.

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