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Vesnalui [34]
4 years ago
9

Suppose you are in an elevator that is moving upward with a constant velocity. A scale inside the elevator shows your weight to

be 600 N.
(a) Does the scale register a value that is greater than, less than, or equal to 600 N during the time when the elevator slows down as it comes to a stop?
(b) What is the reading when the elevator is stopped?
(c) How does the value registered on the scale compare to 600 N during the time when the elevator picks up speed again on its way back down? Give your reasoning
Physics
1 answer:
Ulleksa [173]4 years ago
4 0

Answer:

a) Less than 600 N

b) 600 N

c) Less than 600 N

Explanation:

The value registered by the scale, measures the normal force on the object placed on the scale.

The person on the scale, is acted by two external forces, gravity (which we call weight, and always go downward) and the normal force, an upward force in this case.

The difference between both forces, according Newton’s 2nd law, must be equal to the product of the mass of the object, times the acceleration.

If the elevator moves upward to a constant speed, and then slows down, this means that there exists a net force on the person, producing a acceleration directed downwards on him.

If we take the direction of the acceleration (downward) to be positive, this means that the difference between gravity force and normal force must be positive.

As mg = 600 N (equal to normal force when no net force is present), the normal force in this case must be less than 600 N.

b) If the elevator is stopped, the effect is the same like it were moving at constant speed, so in this case the normal force remains the same: 600 N

c) Assuming that it starts from rest, if it accelerates going downward, and if we take as positive the downward direction, in order to satisfy Newton’s 2nd Law, normal force must be less than 600N so the difference between gravity and normal force remain positive.

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5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v =6t^2 i + 2t j where v is
otez555 [7]

The concept of derivatives and integrals allows to find the results for the questions are the motion of a particle where the speed depends on time are:

       a)the position is:  r = 2 t³ i + t² j

       b) the position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

       c) The acceleration is: a = 12 t i + 2 j

       d) the force is: F = 60 t i + 10 j

       e) the torque is:  τ = 40 t³ k^

       f) tha angular momentum is:  L = 4t³ - 6 t² k^

       g) The kinetic energy is: K = 2 m t² (9t² +1)

       h) The power is:   P = 2m (36 t³ + 2t)

Kinematics studies the movement of bodies, looking for relationships between position, speed and acceleration.

a) They indicate the function of speed.

        v = 6 t² i + 2t j

Ask the function of the position.   The velocity is defined by the variation of the position with respect to time

          v = \frac{dr}{dt}  

          dr = v dt

we substitute and integrate.

        ∫ dr = ∫ (6 t² i + 2t j) dt

        r - 0 = 6 \frac{t^3 }{3} \ \hat i + 2 \frac{t^2}{2 \ \hat j }  

       r = 2 t³ i + t² j

b) The position of the body on the y axis is a parabola and on the x axis it is a cubic function.

c) Acceleration is defined as the change in velocity with time.

           a = \frac{dv}{dt}  

           a = \frac{d}{dt} \ (6t^2 i + 2t j)  

           a = 12 t i + 2 j

d) Newton's second law states that force is proportional to mass times the body's acceleration.

          F = ma

          F = m (12 t i + 2 j)

          F = 5 12 t i + 2 j

          F = 60 t i + 10 j

e) Torque is the vector product of the force and the distance to the origin.

           τ = F x r

The easiest way to write these expressions is to solve for the determinant.

         \tau = \left[\begin{array}{ccc}i&j&k\\F_x&F_y&F_z\\x&y&z\end{array}\right]  

        \tau = \left[\begin{array}{ccc}i&j&k\\60t &10&0\\2t^3 &t^2&0\end{array}\right]  

       τ = (60t t² - 2t³ 10) k

       τ = 40 t³ k ^

f) Angular momentum

        L = r x p

        L =rx (mv)

        L = m (rxv)

The easiest way to write these expressions is to solve for the determinant.

       \left[\begin{array}{ccc}i&j&k\\2t^3 &t^2&0\\6t^2&2t&0\end{array}\right]  

    L = (4t³ - 6 t²) k

 

g) The kinetic energy is

            K = ½ m v²

            K = ½ m (6 t² i + 2t j) ²

            K = m 18 t⁴ + 2t²

            K = 2 m t² (9t² +1)

h) Power is work per unit time

           P = \frac{dW}{dt}dW / dt

The relationship between work and kinetic energy

           W = ΔK

     

          P = 2m \ \frac{d}{dt} ( 9 t^4 + t^2)

          p = 2m (36 t³ + 2t)

In conclusion with the concept of derivatives and integrals we can find the results for the questions are the motion of a particle where the speed depends on time are:

       a) The position is:  r = 2 t³ i + t² j

       b) The position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

       c) The acceleration is: a = 12 t i + 2 j

       d) The force is: F = 60 t i + 10 j

       e) The torque is:  τ = 40 t³ k^

       f) The angular momentum is:  L = 4t³ - 6 t² k^

       g) The kinetic energy is: K = 2 m t² (9t² +1)

       h) The power is:   P = 2m (36 t³ + 2t)

Learn more here:  brainly.com/question/11298125

8 0
2 years ago
Read 2 more answers
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