Plss help me While playing at the park, Katie and her sister take turns going down the slide and playing on the swings.

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

A fire truck drives 3 miles to the east then 1.2 miles to west what is the displacement of the truck

Rudiy27

The answer to the question you have asked is 1.8 miles east.

8 0

3 years ago

Read 2 more answers

What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Paul [167]

Answer:

$\Delta P=1581357.92\ Pa$

Explanation:

Given:

diameter of hose pipe, $D=0.09\ m$

diameter of nozzle, $d=0.03\ m$

volume flow rate, $\dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}$

Now, flow velocity in hose:

$v_h=\frac{\dot V}{\pi.D^2\div 4}$

$v_h=\frac{0.04\times 4}{\pi\times 0.09^2}$

$v_h=6.2876\ m.s^{-1}$

Now, flow velocity in nozzle:

$v_n=\frac{\dot V}{\pi.d^2\div 4}$

$v_n=\frac{0.04\times 4}{\pi\times 0.03^2}$

$v_n=56.5884\ m.s^{-1}$

We know the Bernoulli's equation:

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2$

when the two points are at same height then the eq. becomes

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}$

$\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}$

$\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}$

$\Delta P=1581357.92\ Pa$

8 0

3 years ago

What is a resistor? A battery A light bulb A switch

melomori [17]

The answer is a battery

8 0

3 years ago

a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s . part a calculate its total kinet

erma4kov [3.2K]

49 J is the total kinetic energy. If a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s. Kinetic energy is the energy an bowling ball has because of its motion.

Given: m = 7.3 Kg ; r = 9.4 cm = 0.094 m ; v = 3.1 m

Now total kinetic energy in this case is given by KE = Kinetic energy due to rotation + Kinetic energy due to translation

i,e KE = 1/2*m*v2 + 1/2*I*ω2 where I is the moment of inertia of the bowling ball about it's center and ω is the angular velocity

Now for pure rotation (without slipping) v = rω

also for the ball (solid sphere) I = 2/5*m*r2

Hence our kinetic energy becomes

KE = 1/2*m*v2 + 1/5*m*v2 = 7/10*m*v2

so KE = 0.7*7.3*(3.1)2 = 49.10 J = 49 J

Learn more about kinetic energy here

brainly.com/question/12669551

#SPJ4

5 0

2 years ago