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Vanyuwa [196]
3 years ago
11

A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle

ss surface below. Suppose you pull the mass so that the spring is stretched out (initial state) and then you release it, so that the mass starts moving towards the spring is unstretched position (final state). The impulse imparted on the spring-mass system by the force that the wall exerts on the spring is zero, since the wall does not move during this process.
Required:
What total percentage of the period does the mass lie in these regions?
Physics
1 answer:
Radda [10]3 years ago
6 0

Answer:

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

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What type of system allows energy but not matter to enter and exit?
Nastasia [14]

Answer:

closed

Explanation:

A closed system allows energy (usually heat) to be exchanged but not matter.

Your welcome

5 0
3 years ago
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
pochemuha

Since g is constant,  the force the escaping gas exerts on the rocket will be 10.4 N

<h3>What is Escape Velocity ?</h3>

This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.

Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.

To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters

  • Mass (m) of the rocket 0.25 Kg
  • Initial velocity u = 15 m/s
  • Final Velocity v = 40 m/s
  • Time t = 0.6s
  • Gravitational field intensity g = 9.8N/kg

The force the gas exerts of the rocket = The force on the rocket

The rate change in momentum of the rocket = force applied

F = ma

F = m(v - u)/t

F = 0.25 x (40 - 15)/0.6

F = 0.25 x 41.667

F = 10.42 N

Since g is constant,  the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.

Learn more about Escape Velocity here: brainly.com/question/13726115

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7 0
2 years ago
14. What is the mass of a 250 N bag of groceries?
Veseljchak [2.6K]

Answer:

mass = 25.5kg

Explanation:

The gravitational force equivalent, or, more commonly, g-force, is a measurement of the type of force per unit mass – typically acceleration – that causes a perception of weight, with a g-force of 1 g equal to the value of gravitational acceleration on Earth, g, of about 9.8 m/s².

Mass, m = \frac{Force}{acceleration due to gravity}

mass,m=\frac{F}{g} = \frac{250}{9.8}= 25.5 kg

8 0
3 years ago
Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?
cluponka [151]

Answer:

Torque, \tau=0i+0j-70k

Explanation:

It is given that,

Force acting on the particle, F=-10j\ N

Position of the particle, r=(7i+5j)\ m

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

\tau=r\times F

\tau=(7i+5j)\times (-10j)

The cross product of vectors is given by :

\tau=\begin{pmatrix}0&0&-70\end{pmatrix}

or

\tau=0i+0j-70k

So, the torque on the particle about the origin 0i+0j-70k. Hence, this is the required solution.

6 0
3 years ago
What does the graph to the right represent?
Nastasia [14]
C the thermal equilibrium

4 0
4 years ago
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