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allsm [11]
3 years ago
11

Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H

e tells you that the force is always pointed away a definite point in space, which we can call the force center. The magnitude of the force turns out to be proportional to B/r3, where r is the distance from the force center to any other point. Your friend says that it has been determined that the constant of proportionality has been determined to be B= 2 (in units to be determined later), so that the magnitude of the force on a particle (in newtons) can be written as 2r^3, when the particle is at a distance r from the force center.
Required:
Write an expression of potential energy.
Physics
1 answer:
kirill [66]3 years ago
5 0

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

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A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots
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(a). The car's average velocity between t = 1.0s to t = 1.5s will be - 1\;m/s

(b). The car's acceleration at t = 1.5s will be - 0.4\;m/s^{2}

(c). Car's speed is increasing with time.

We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.

We have to determine -

  • The car's average velocity (in m/s) in the interval between -

        t = 1.0 s  to  t = 1.5 s.

  • The car's acceleration at t = 1.5 s.
  • Determining whether car's speed increasing or decreasing with time.

<h3>What is Acceleration?</h3>

The rate of change of velocity with respect to time is called Acceleration. Mathematically -

$a=\frac{dv}{dt}

According to the question, we have the following data for the Car -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m

PART - A

The car's average velocity between t = 1.0s to t = 1.5s will be -

$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s

PART - B

Velocity at t = 1.5 s will be -

$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s

The car's acceleration at t = 1.5s will be -

$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}

PART - C

Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.

[ The following data was missing in your answer. The complete question would include this data also -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m ]

To solve more questions on Kinematics, visit the link below-

brainly.com/question/17272824

#SPJ2

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An initially motionless test car is accelerated uniformly to 120 km/h in 8.28 s before striking a simulated deer. The car is in
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Answer:

The value is   a =  4.0 \ m/s^2

Explanation:

From the question we are told that  

  The velocity  is  a =  120 \  km/h =  [tex]\frac{120 *  1000}{3600} =  33.3 \  m/s[/tex]

  The time taken is t  =  8.28 \  s

    The  time taken for contact is  t_c  =  0.815 \  s

     The  velocity of the of the car after contact is v_c =  71.0 \  km/h = [tex]\frac{71 *1000}{3600} =  19.7 2 \  m/s[/tex]

From the equation of kinematics we have that  

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Here   u =  0 \ m/s  since the car is initially motionless

=>    33.3 =  0  + a *  8.28

=>    a =  4.0 \ m/s^2

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Answer:

C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.

Explanation:

When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.

If it were a series connection, then the current would have been the same, but the voltage drop were another story.

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