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pishuonlain [190]
3 years ago
12

A block of mass m is attached to the end of a spring (spring stiffness constant k ). The mass is given an initial displacement x

0 from equilibrium, and an initial speed v 0 . Ignoring friction and the mass of the spring, use energy methods to find its maximum speed. Express your answer in terms of some or all of the variables v0, k , m , and x 0 .
Physics
1 answer:
LiRa [457]3 years ago
7 0

Answer:

vvbbbjjjjjjhddyibcxxuio000

Explanation:

gg

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a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio
Likurg_2 [28]

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

  • mass of the ball, m₁ = 0.8 kg
  • speed of the ball, u₁ = 2.5 m/s
  • mass of the object at rest, m₂ = 2.5 kg
  • final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁  +  2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

5 0
2 years ago
Carbon is allowed to diffuse through a steel plate 11 mm thick. The concentrations of carbon at the two faces are 0.88 and 0.41
Serhud [2]

Answer:

The temperature is 2584.5 K

Explanation:

Given:

Activation energy Q = 80000 \frac{J}{mol}

Preexponential D= 6.2 \times 10^{-7} \frac{m^{2} }{s}

Diffusion flux J = 6.4 \times 10^{-10} \frac{kg}{m^{2} s}

Thickness of plate \Delta x =  11 \times 10^{-3} m

Concentration of carbon at two faces \Delta C = (0.88 - 0.41 ) = 0.47 \frac{kg}{m^{3} }

From the formula of temperature in terms of diffusion flux,

  T = (\frac{Q}{R} ) \frac{1}{\ln (\frac{D\Delta C}{J\Delta x} )}

Where R = 8.314 \frac{J}{mol.K} ( gas constant )

Put the values and find the temperature,

  T = (\frac{80000}{8.314} ) \frac{1}{\ln (\frac{6.2 \times 10^{-7} \times 0.47 }{6.4 \times 10^{-10}\times 11 \times 10^{-3} } )}

  T = 2584.5 K

Therefore, the temperature is 2584.5 K

8 0
3 years ago
An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only
topjm [15]

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron  )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² =  1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities  are  inversely proportional to square root of their masses .

4 0
3 years ago
Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin
Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

       θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

     w₀B = ½ w₀

     αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

      t´ = t - t_1

l

et's write the equation of particle B

      θ = θ₀ + w₀B t´ + ½ αB t´2

replace

     θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

     θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0
2 years ago
According to the _______ the amount of energy in the universe doesn't change.
balandron [24]
The answer is B, Law of Kinetic Energy
6 0
3 years ago
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