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stellarik [79]
3 years ago
14

Biker A is cruising at a speed of 10.0 m/s when she passes biker B who is at rest at the origin of the coordinate system. Biker

B immediately accelerates when he is passed, catching up with biker A after 1.00 minute.
(a) Draw graphs of position, velocity, and acceleration of both bikers as a function of time.
(b) What is the acceleration of biker B, assuming that it is constant?
(c) At what distance from the origin do the two bikers meet?
(d) What is the average speed of biker B during the first 30.0 seconds?

Physics
1 answer:
Burka [1]3 years ago
3 0

Answer: attached

Explanation:

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A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

        x = 0.135 cm

Given parameters

  • The mass m = 220 g = 0.220 kg
  • The spring cosntnate3 k = 7.0 N / m
  • Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

  • The position where the kinetic and potential energy are equal

 

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

          x = A cos wt + fi

          w² = \frac{k}{m}

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

       v = \frac{dx}{dt}

let's evaluate

       v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

       0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

        x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

        K = U

        ½ m v² = m g x

       

we substitute

        ½ A² w² sin² wt = g A cos wt

        sin² wt = \frac{2g}{A}  cos wt

let's calculate

      w = \sqrt{\frac{7}{0.220} }  

      w = 5.64 rad / s

      sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

      sin² 5.64t = 376.92 cos 5.64 t

      1 - cos² 5.64t = 376.92 cos 5.64t

      cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

       x = cos 5.64t

      x²- 376.92 x - 1 = 0

      x = 0.026

      cos 5.64t = 0.026

   

Let's find the displacement for this time

       x = 5.2 10-2 0.026

       x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

        x = 0.135 cm

Learn more here: brainly.com/question/15707891

8 0
3 years ago
A beam of yellow light is made to pass through two slits that are 3.0 x 10−3 meters apart. On a screen 2.0 meters away from the
lakkis [162]

Answer:

585 nm

Explanation:

The formula that gives the position of the m-th maximum (bright fringe) relative to the central maximum in the interference pattern produced by diffraction from double slit is:

y=\frac{m\lambda D}{d}\Delta y =\frac{m\lambda D}{d}

where

m is the order of the maximum

\lambda is the wavelength

D is the distance of the screen from the slits

d is the separation between the slits

The distance between two consecutive bright fringes therefore is given by:

\Delta y = \frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}=\frac{\lambda D}{d}

In this problem we have:

\Delta y = 3.9\cdot 10^{-4} m (distance between two bright fringes)

D = 2.0 m (distance of the screen)

d = 3.0 x 10−3 m (separation between the slits)

Solving for \lambda, we find the wavelength:

\lambda=\frac{\Delta y d}{D}=\frac{(3.9\cdot 10^{-4})(3.0\cdot 10^{-3})}{2.0}=5.85\cdot 10^{-7} m = 585 nm

4 0
3 years ago
Which of the following describes the time over which a periodic wave repeats?
sineoko [7]

Answer:

b

Explanation:

4 0
4 years ago
One of the solid reactants was treated in a coffee grinder before adding to
Nikitich [7]

Answer:

C. Surface area

Explanation:

The rate of  chemical reaction depends on various factors such as:

  • concentration and pressure
  • nature of reactants
  • temperature
  • surface area
  • presence of catalyst, etc.

Effect of surface area of reactants: the rate of a chemical reaction can be increased by increasing the the area of contact of the reacting substances. This is especially important when one or more of the reactants are solids., because only the particles of the solid that are exposed are able to take part in the reaction at each instant of time. Therefore, the greater the surface area of the solid reactant particles the faster the reaction.

The surface area of solid reactants can be increased by grinding or pelletizing, thus allowing for a greater contact between the reacting particles,

The instance in which one of the solid reactants was treated in a coffee grinder before adding to the reaction container is one way of increasing the surface area of a reactant.

7 0
3 years ago
A 2.7-kg block slides with a speed of 1.1 m/s on a frictionless, horizontal surface until it encounters a spring.
IceJOKER [234]

Answer

given,

mass of block = 2.7 kg

speed of block = 1.1 m/s

a) spring compressed to distance of 6 cm or 0.06 m

  using conservation of energy

kinetic energy is converted into spring energy

\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2

k =\dfrac{ mv^2}{x^2}

k =\dfrac{2.7\times 1.1^2}{0.06^2}

k= 907.5 N/m

b) again using conservation of energy

kinetic energy is converted into spring energy

\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2

v =\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{907.5\times 0.015^2}{2.7}}

v=\sqrt{0.075625}

v= 0.275 m/s

8 0
3 years ago
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