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Setler [38]
3 years ago
9

If a particle's position is given by x = 4 − 12 t + 3 t^ 2 (where t is in seconds and x is in meters), what is its velocity at s

t = 1 s ? b) Is it moving in the positive or negative direction of x just then? c) What is its speed just then? d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculations.) e) Is there ever an instant when the velocity is zero? If so, give the time t ; if not, answer no. f) Is there a time after t = 3 s when the particle is moving in the negative direction of x ? If so, give the time t ; if not, answer no.
Physics
1 answer:
sweet [91]3 years ago
6 0

a) -6 m/s

The particle's position is given by

x=3t^2-12t+4

where t is the time.

The velocity of the particle can be found by calculating the derivative of the particle's position with respect to the time:

v=x'(t)=3(2t)-12=6t-12

So, the velocity at instant t=1 can be found by substituting t=1 in this formula:

v(1 s)=6(1)-12=-6 m/s

b) Negative direction

At instant t=1 second, the particle's velocity is

v=-6 m/s

we see that this velocity is negative: therefore, this means that the particle is moving along the negative x-direction.

c) 6 m/s

The particle's speed is just equal to the magnitude of the velocity:

speed = |v|

Since the speed is a scalar and velocity is a vector.

Since the velocity is

v=-6 m/s

Then, the speed of the particle is

speed = |-6 m/s| = 6 m/s

d) Increasing

The speed is the magnitude of the velocity, so we can write

speed = |v| = |6t-12|

We notice that at t = 1 s, for little increments of t the term (6t) is increasing in magnitude, while the constant term -12 remains constant. This means that the whole function

|6t-12|

is increasing at t=1 s, so the speed is increasing.

e) yes, at t = 2 s

In order to find an instant in which the velocity is zero, we just need to set v(t)=0 in the equation:

6t-12 =0

and solving the equation for t, we find

6t=12\\t=\frac{12}{6}=2

So, at t = 2 seconds the velocity is zero.

f)  No

The particle is moving in the negative direction of x when the velocity vector v(t) is negative.

The velocity vector is

v(t) = 6t -12

we see that at t=3 s, it is equal to

v(3) = 6(3)-12 = 18-12=+6

so it is positive.

For every time t > 3 s, we notice that the term (6t) in the formula is increasing and always positive, and since it is always larger than the constant negative term (-12), we can conclude that the velocity is always positive for every time > 3 seconds.

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Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
2 years ago
An 80 N force causes a spring to compress 0.15 m. What is the spring constant? What is the potential energy of the spring?
Anni [7]
Force = -kx
80N=0.15m * -k
K=-80/0.15=533.333. Spring constant
Energy=1/2kx^2
1/2*(-80/0.15)*80^2=Energy
3 0
3 years ago
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
____ is undeniably accepted by scientists all over the world as the primary language of science.
faust18 [17]

Answer:

Latin

Explanation:

In order for the scientists to have a common and official name for a particular thing that can be understood by every scientist in the world, a single language has been established for the purpose. The language chosen is the Latin language. The official scientific names are given in this language, so it is a necessity for the scientists to know and understand this language. The terms that are commonly used are regional, and they come in many different languages, which is why this language has been chosen. Occasionally, the ancient Greek language is used as well, though much less than the Latin.

6 0
3 years ago
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How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


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3 years ago
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