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masya89 [10]
3 years ago
13

A 40.1 g object is attached to a horizontal spring with a spring constant of 11.9 N/m and released from rest with an amplitude o

f 24.7 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
Physics
1 answer:
Arlecino [84]3 years ago
3 0

Answer: v = 3.684 m/s

Explanation: The angular frequency (ω) of a loaded spring is given as

ω = √k/m

Where ω = angular frequency, k =spring constant = 11.9 N/m, m = mass of object = 40.1 g = 0.0401 kg.

The velocity of a simple harmonic motion is defied as

v = ω√A² - x²

Where A = Amplitude = 24.7cm = 0.247m and x = displacement.

For our question, we where asked to find velocity at half way, at half way, x = A/2

Hence at half way, x = 0.247/2 = 0.1235 m.

We need to get the value of angular frequency first.

ω = √(11.9/0.0401)

ω = √296.758

ω = 17.22 rad/s.

Then the velocity is

v = 17.22 √0.247² - 0.1235²

v = 17.22 √0.061009 - 0.01525225

v = 17.22 √0.04575675

v = 17.22 × 0.2139

v = 3.684 m/s

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VikaD [51]

Answer: 580 x 10^-3 J

Explanation:

0.6mm is 0.6/1000 = 600*10^-6 m

The plate area is .17*.17 = 28.9*10^-3 m^2

Air:

The voltage that can be sustained by 0.60 mm of air dielectric is:

V = 3.0*10^6* 600*10^-6 = 1800 V

The capacitance is:

C = ε*A/d = 8.854*10^-12 * 28.9*10^-3/600*10^-6 = 426*10^-12 F = 426 pF

The energy stored in a capacitor is:

E = (1/2)*C*V^2 = (1/2)*426*10^-12*(1800)^2 = 691*10^-6 J

Teflon:

The voltage is:

V = 60*10^6* 600*10^-6 = 36*10^3 = 36 kV

According to the listed reference, the relative dielectric constant for teflon is 2.1, this figure multiplies the "ε" of free space.

The capacitance is:

C = ε*A/d = 2.1*8.854*10^-12 * 28.9*10^-3/600*10^-6 = 896*10^-12 F = 896 pF

It would have been easier to note that the capacitance is 2.1 times the air-dielectric case.

The maximum energy stored is:

E = (1/2)*C*V^2 = (1/2)* 896*10^-12* (36*10^3)^2 = 580*10^-3 J

8 0
3 years ago
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mario62 [17]

Answer:

A. The core makes up the majority of Earth's volume.

Explanation:

Although the core and mantle are about equal in thickness, the core actually forms only 15% of earth's volume whereas the mantle takes up 84%

8 0
3 years ago
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
dimulka [17.4K]

Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      W =  E_h  -E_c

substituting values

       W = 63000 J

So

    P =  \frac{63000}{3600}

    P =  17.5 W

The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

         \eta_c  =  1 - \frac{322}{1250}

         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

7 0
3 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
  • This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
5 0
3 years ago
What is the volume of a cube with length = 3 centimeters, width = 3
suter [353]

Answer:

D. 27 cubic centimeters

Explanation:

l*w*h

3*3*3= 27

4 0
3 years ago
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