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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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Answer:

i dont know

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6 0
2 years ago
A spring that is compressed 14.5 cm from its equilibrium position stores 2.99 J of potential energy. Determine the spring consta
strojnjashka [21]

Answer:

284.4233 N/m

Explanation:

k = Spring constant

x = Compression of spring = 14.5 cm

U = Potential energy = 2.99 J

The potential energy of a spring is given by

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

Rearranging to get the value of k

\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 2.99}{0.145^2}\\\Rightarrow k=284.4233\ N/m

The spring constant is 284.4233 N/m

7 0
3 years ago
Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0
3 years ago
Read 2 more answers
What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

4 0
3 years ago
A wedge with a mechanical advantage of 0.78 is used to raise a house corner from its foundation. If the output force is 7500 N,
Zigmanuir [339]

Answer:

5850

Explanation:

MA = Force out/ Force in

.78 = out/ 7500

.78 times 7500 = input

6 0
3 years ago
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