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lara [203]
4 years ago
7

You’re driving your car towards an intersection. A Porsche is stopped at the red light. You’re traveling at 36 km/h (10 m/s). As

you are 15 m from the light, the light turns green, and the Porsche accelerates from rest at 3 m/s2. You continue at constant speed. a. How far from the stop line do you pass the Porsche? At what time, measured from when the light turned green, do you pass the Porsche? b. As the Porsche keeps accelerating, it eventually catches up to you again. How far from the stop line does it pass you? At what time, measured from when the light turned green, does it pass you? c. If a Boston police officer happens to get you and the Porsche on a radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/h.
Physics
1 answer:
34kurt4 years ago
6 0

Your position at time t, relative to the stop line:

x_1=-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t

The Porsche's position:

x_2=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2

a. You pass the Porsche immediately after the time it takes for x_1=x_2:

-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=2.3\,\rm s

at which point you both will have traveled 7.8 m from the stop line.

b. The equation in part (a) has two solutions. The Porsche passes you at the second solution of about t=4.4\,\rm s, at which point you both will have traveled 29 m.

c. At time t, the Porsche is moving at velocity

v=\left(3\dfrac{\rm m}{\mathrm s^2}\right)t

so that at the moment it passes you, its speed is 13 m/s, which is about 46.8 km/h and below the speed limit, so neither of you will be pulled over.

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marysya [2.9K]

Answer:

is 3 and 2

Explanation: the firth one is 3 and the 2

6 0
3 years ago
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A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A
monitta

Answer:

induced EMF = 240 V

and by the lenz's law  direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L \frac{dI}{dt}    ...................1

so E = - L \frac{I2 - I1}{dt}

put here value we get

E = - 8 × 10^{-3} \frac{10 - 4}{0.2*10^{-3}}

E = -40 ×  6

E = -240

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induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

5 0
3 years ago
The Doppler shift can be observed when ____________. Check all that apply. Group of answer choices the source of sound moves tow
castortr0y [4]

Answer:

the source of sound moves towards an observe

Explanation:

The Doppler effect is related to waves such as sound or light. the effect causes an increase or decrease in the frequency of sound light or other waves when the souces either move towards or away from the observer. For example the siren of the train to a person on the platform, the redshift seen by astronomers.

Therefore, The Doppler shift can be observed when the source of sound moves towards an observer From a place closer to the observer than the last wave's crest, each consecutive wave crest is sent. Each wave therefore, takes a little less time than the preceding wave to reach the observer.

6 0
3 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
Help not to sure with this one need help plz asap​
DiKsa [7]

A is the answer for the problem

4 0
3 years ago
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