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Tpy6a [65]
3 years ago
6

The rotating nozzle sprays a large circular area and turns with the constant angular rate theta overscript dot endscripts = 2.2

rad/s. particles of water move along the tube at the constant rate l overscript dot endscripts = 1.8 m/s relative to the tube. find the magnitudes of the velocity v and acceleration a of a water particle p when the distance l = 0.85 m if the angle beta = 64°.

Physics
1 answer:
densk [106]3 years ago
3 0
Refer to the figure shown below.

ω = 2.2 rad/s, the constant angular velocity of rotation
u = 1.8 m/s, the radial velocity of a water particle, P
v = tangential velocity of P at radius r = 0.85 m from the sprinkler
P is at angle 64° from the horizontal axis.

Calculate the tangential velocity of P.
v = rω = (0.85 m)*(2.2 rad/s) = 1.87 m/s

Calculate the centripetal acceleration.
a = v²/r =(1.87 m/s)²/(0.85 m) = 4.114 m/s².
This acceleration is directed toward the sprinkler.

The magnitude of the resultant particle velocity is
V = √(v² + u²)
    = √(1.87² + 1.8²)
    = 2.596 m/s

Because tan⁻¹ (v/u) = tan⁻¹ (187/1.8) = 46°, the resultant particle velocity is
64 + 46 = 110° measured counterclocwise from the positive horizontal x-axis.

Answer:
The velocity of particle P is 2.6 m/s at an angle of 110° measured counterclockwise from the positive x-axis.
The acceleration of particle P is 4.1 m/s², and it is directed toward the sprinkler.

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The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

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(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

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(c) Yes

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V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

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So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

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V_L=\frac{V}{2}

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And solving the equation for t, we find the time t at which the two voltages are equal:

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V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

V_L = 0

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