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Fudgin [204]
3 years ago
14

It is known that the population mean for the verbal section of the SAT is 500 with a standard deviation of 100. In 2006, a sampl

e of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511. The point estimate of the mean for this group is _____ and the 95 percent confidence interval for this group is _____.
Physics
1 answer:
kvasek [131]3 years ago
4 0

Answer

given,

SAT is 500 with a standard deviation of 100.

a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.

sample mean = \dfrac{standard\ deviation}{\sqrt{n}}

                      = \dfrac{100}{\sqrt{400}}

                      = 5

95% confidence level is achieved within +/- 1.960 standard deviations.

1.960 standard deviations  x  5 is equal to +/- 9.8

confidence interval = 511 - 9.8   ---  511 + 9.8

                                = 501.2-----520.8

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Anuta_ua [19.1K]

Answer:the rate of change in velocity

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While velocity is the distance traveled in a given direction/time taken.the unit is m/s

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Why would we expect protogalactic clouds with relatively high density to form an elliptical galaxy rather than a spiral galaxy?
Svetlanka [38]

The higher density allows the protogalactic clouds to cool faster and form an elliptical galaxy rather than a spiral galaxy.

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Since there had been no previous star formation to create other elements, protogalaxies would have been made up almost entirely of hydrogen and helium. The hydrogen would bond to form H2 molecules, with some exceptions. This would change as star formation began and produced more elements through the process of nuclear fusion.

Mechanics

Once a protogalaxy begins to form, all particles bound by its gravity begin to free fall towards it. The time taken for this free-fall to conclude can be approximated using the free-fall equations. Most galaxies have completed this free-fall stage to become stable elliptical or disk galaxies, the disks taking longer to fully form. The formation of galaxy clusters takes much longer and is still in progress now.

This stage is also where galaxies acquire most of their angular momentum. A protogalaxy acquires this due to gravitational influence from neighbouring dense clumps in the early universe, and the further the gas is away from the centre, the more spin it gets.

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7 0
2 years ago
A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces up
Serjik [45]

Answer:

0.906

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So the efficiency of the water turbine is the ratio of output power over input power:

\frac{234}{258.307} = 0.906

3 0
3 years ago
Velocity is defined as:
gogolik [260]

Velocity is defined as a change in position.

6 0
2 years ago
A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu
lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

       F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)

b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

d)

  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
  • So, we can use the following kinematic equation in order to find the displacement before coming to rest:

        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x  (7)

  • Since the puck comes to a stop, vf =0.
  • Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

       \Delta x  = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m  (8)

e)

  • The total work done by the friction force on the object , can be obtained in several ways.
  • One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
  • Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2}   = -12.4 J  (9)

f)

  • By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
  • P_{Avg} = \frac{\Delta E}{\Delta t}  (10)
  • If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

       P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)

g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

7 0
3 years ago
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