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2 years ago

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Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving

at speed vi on a frictionless surface while attached to an ideal, horizontally fixed spring with a force constant of ki . Let x denote the displacement of the block from its equilibrium position. Order the systems from largest total mechanical energy to smallest.
a. m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
b. m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
c. m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
d. m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m

1 answer:

Rzqust [24]2 years ago

4 0

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

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8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

2 years ago

An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around whic

Tpy6a [65]

Answer:

Part a)

$a = 0.056 m/s^2$

Part b)

$L = 7.85 m$

Explanation:

Part a)

Angular speed of the pedal is changing from 60 rpm to 90 rpm in 10 s

so the angular acceleration is given as

$\alpha = \frac{\omega_2 - \omega_1}{\Delta t}$

so we will have

$\alpha = \frac{2\pi(\frac{90}{60}) - 2\pi(\frac{60}{60})}{10}$

$\alpha = 0.314 rad/s^2$

now the tangential acceleration of the pedal is given as

$a = r \alpha$

$a = 0.18 \times 0.314$

$a = 0.056 m/s^2$

Part b)

Total angular displacement made by the sprocket in the interval of 10 s is given as

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{2\pi (\frac{90}{60}) + 2\pi (\frac{60}{60})}{2}(10)$

$\theta = 78.5 radian$

now length of the chain passing over it is given as

$L = R\theta$

$L = 0.10 \times 78.5$

$L = 7.85 m$

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2 years ago

vector A makes equal angles with x,y and z axis. value of its components (in terms of magnitude of vector A will be?

kiruha [24]

X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared

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1 year ago

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A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The fina

Natali [406]

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

$\Delta U = \frac{3}{2}nR\Delta T$

now for isothermal process since temperature change is zero

so change in internal energy must be ZERO

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2 years ago

A free-falling golf ball strikes the ground and exerts a force on it. Which sentences are true about this situation? A golf ball

Harlamova29_29 [7]

Answer:

The ground exerts an equal force on the golf ball

Explanation:

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

In this problem, object A is the golf ball while object B is the ground, so we can say that:

- the golf ball exerts a force on the ground

- the ground exerts an equal and opposite force on the golf ball

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1 year ago

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