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kvasek [131]
3 years ago
10

An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from

its original length when it reaches equilibrium. The mass is then lifted up a distance L=0.0325 m from the equilibrium position and released. What is the kinetic energy of the mass at the instant it passes back through the equilibrium position?
Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

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Answer:

Part a)

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As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery

So here we can say that charge on the plates will remain conserved

So we will have

Q = kC(2.1)

now dielectric is removed between the plates of capacitor

so new potential difference between the plates

V' = \frac{Q}{C'}

V' = \frac{kC(2.1)}{C}

V' = 3.7 \times 2.1

V' = 7.77 Volts

Part b)

Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates

So again charge is same so potential difference is given as

V" = \frac{Q}{k'C}

0.59 V = \frac{kCV}{k'C}

0.59 = \frac{3.7}{k'}

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5 0
3 years ago
Picture a long, straight corridor running east-west, with a water fountain located somewhere along it. Starting from the west en
Brut [27]

Answer:

The corridor's distance is "90 m".

Explanation:

  • She heads in the east directions but creates the first pause, meaning she crosses the distance 'x' in step 1.
  • Now, provided that perhaps the distance by her to another fountain or waterfall just after the first stop is twice as far away she traveled.
  • Because she moved the distance of 'x,' then, therefore, her distance towards the fountain of '2x.' She casually strolls and once again pauses 60 m beyond her stop.
  • The gap about her to the waterfall during that time approximately twice the distance and her to the eastern end of the hallway.
  • Assume her gap from either the east end of the platform seems to be 'y' at either the second stop, after which '2y' may become the distance between the 2nd pause and the waterfall.

Now,

⇒  2x + 2y = 60

⇒  x + y = 30

The total distance of the corridor will be:

=  x + 2x + 2y + y

=  3\times (x + y)

=  3\times 30

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4 0
3 years ago
Car A (mass 1100 kg) is stopped at a traffic light when it is rear­ended by car B(mass 1400 kg). Both cars then slide with locke
viva [34]

Answer:

Part a)

v_a = 3.94 m/s

Part b)

v_b = 3.35 m/s

Part C)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

a = -\frac{F_f}{m}

a = -\frac{\mu mg}{m}

a = - \mu g

now we will have

v_f^2 - v_i^2 = 2ad

0 - v_i^2 = 2(-\mu g)(6.1)

v_a = \sqrt{(2(0.13)(9.81)(6.1)}

v_a = 3.94 m/s

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

v_b = \sqrt{2(0.13)(9.81)(4.4)}

v_b = 3.35 m/s

Part C)

By momentum conservation we will have

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1400 v_b = 1100(3.94) + 1400(3.35)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0
3 years ago
I WILL MARK BRAINLIEST IF CORRECT!!!
julsineya [31]

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N I C E - D A Y!

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Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is
goblinko [34]

Answer:

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= 630×10^{-9} m × 3×3m/ 45×10^{-3} m

= 1.26×10^{-4}m

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