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Nadya [2.5K]
3 years ago
12

50 POINTS

Physics
2 answers:
sweet-ann [11.9K]3 years ago
7 0

Answer:

The acceleration is constant and positive

Explanation:

The straight line indicates that the acceleration is constant, while the positive slope indicates that the line is positive.

yawa3891 [41]3 years ago
7 0

Answer:

^^^

Explanation:

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A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
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3 years ago
What is number 14 please tell me
UNO [17]
<span>a thin fibrous cartilage between the surfaces of some joints, e.g., the knee.</span>
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3 years ago
A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a
Art [367]

Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

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If the cross-section of a wire of fixed length is doubled, how does the resistance of that wire change? (this is for studying fo
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If the cross-section of a wire of fixed length is doubled,  the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>
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