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°.
1 answer:
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.
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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Answer:
At t = 15.0 s the magnitude of the velocity is 58.31 m/s
Explanation:
The given parameters are;
V₀ₓ = 30 m/s
= 100 m/s
The time of flight of the projectile = 25 s
For projectile motion;
Vₓ = V₀ × cos(θ₀)
The magnitude of the velocity V = √(V₀ₓ² + ²)
We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s
cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109
θ₀ = cos⁻¹(3/√109) = 73.3°
The components of the velocity after time t is given by the relations;
Vₓ = V₀ × cos(θ₀) = 30 m/s
= V₀ × sin(θ₀) - g×t
When = 0, we have;
0 = V₀ × sin(θ₀) - g×t
g×t = V₀ × sin(θ₀) = 10·√109×0.958 = 100 m/s
t = 100/g = 100/10 = 10 s
The time to reach maximum height = 10 s
At 15.0 seconds, we have;
= V₀ × sin(θ₀) - g×t = 10·√109×0.958 - 10×15 = -50 m/s
Therefore, the projectile is returning at 50 m/s
The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.
Answer:
The power factor of the circuit is 0.99
Explanation:
Given;
resistance of the resistor, R = 65.2 ohms
capacitance of the capacitor, C = 2.26 μF = 2.26 x 10⁻⁶ F
inductance, L = 2.08 mH = 2.08 x 10⁻³ H
frequency of the AC, f = 2400 Hz
The angular frequency is given by;
ω = 2πf
ω = 2π(2400) = 15081.6 rad/s
The inductive reactance is given by;
XL = ωL
XL = (15081.6 x 2.08 x 10⁻³)
XL = 31.37 ohms
The capacitive reactance is given by;
The phase difference is given by;
The power factor is given by;
CosФ = Cos(1.78) = 0.99
Therefore, the power factor of the circuit is 0.99