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Mila [183]
3 years ago
15

A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What wi

ll be the compression if the same block collides with the spring at a speed of 2v?
Physics
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

4.0 cm

Explanation:

For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v

Since m and k are constant since its the same spring x ∝ v

If our speed is now v₁ = 2v, our compression is x₁

x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x

x₁ = 2x

Since x = 2.0 cm, our compression for speed = 2v is

x₁ = 2(2.0) = 4.0 cm

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Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
3 years ago
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Based on illustrations of magnetic field lines, where could an object be placed so it would not experience a magnetic force
tiny-mole [99]

<u>Halfway</u><u> between the like poles of two magnets, because the field lines bend away and do not enter this area.</u>

How does a magnetic field diagram show where the field is strongest?

  • The magnetic field lines do not ever cross.
  • The lines include arrowheads to indicate the direction of the force exerted by a magnetic north pole.
  • The closer the lines are to the poles, the stronger the magnetic field (thus the magnetic field from a bar magnet is highest closest to the poles).

Where is magnetic field the strongest and weakest on a magnet?

  • The bar magnet's magnetic field is strongest at its core and weakest between its two poles.
  • The magnetic field lines are densest immediately outside the bar magnet and least dense in the core.

Which two locations on the magnet would have the greatest attractive forces?

  • Inside the magnet itself, the field lines run from the south pole to the north pole.
  • The magnetic field is strongest in areas of greatest density of magnetic field lines, or areas of the greatest magnetic flux density.

Learn more about magnetic field

brainly.com/question/11514007

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4 0
2 years ago
In which situation would an object weigh the LEAST? (assume all the objects have the same mass)
inessss [21]

Answer:

An object on the moon would weigh the LEAST among these. So correct answer is B.

Explanation:

  • Weight of an object on any place is given by:

W = Mass * Acceleration due to gravity(g)

  • It means when masses of different objects those are in different places are same, the weight of  those objects depends upon the 'g' of that particular place.
  • As we know, acceleration due to gravity on surface of moon (g') is 6 times weaker than the acceleration on surface of earth (g), which is due to the large M/R^2 of the earth than the moon.

i.e. g' = g/6 so W' = W/6

  • And in the space between the two, the object is weightless.
8 0
3 years ago
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A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the e
Makovka662 [10]

Answer:

 y = -19.2 sin (23.15t) cm

Explanation:

The spring mass system is an oscillatory movement that is described by the equation

      y = yo cos (wt + φ)

Let's look for the terms of this equation the amplitude I

     y₀ = 19.2 cm

Angular velocity is

     w = √ (k / m)

     w = √ (245 / 0.457

     w = 23.15 rad / s

The φ phase is determined for the initial condition   t = 0 s ,  the velocity is negative v (0) = -vo

The speed of the equation is obtained by the derivative with respect to time

     v = dy / dt

     v = - y₀ w sin (wt + φ)

For t = 0

     -vo = -yo w sin φ

The angular and linear velocity are related v = w r

      v₀ = w r₀

      v₀ = v₀ sinφ

      sinφ = 1

      φ = sin⁻¹ 1

      φ = π / 4    rad

Let's build the equation

      y = 19.2 cos (23.15 t + π/ 4)

Let's use the trigonometric ratio π/ 4 = 90º

      Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a

       y = -19.2 sin (23.15t) cm

8 0
3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
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