When Aditya pushes on Rachel and her bicycle, they accelerate at 0.22 m/s/s. If Aditya pushes on Rachel and her bicycle with twi
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Refer to the figure shown below, which is based on the given figure.
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
![4.9[ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ]^{2} - v_{0} sin(\theta - \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ] - X sin \alpha = 0](https://tex.z-dn.net/?f=4.9%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%20%5D%5E%7B2%7D%20-%20v_%7B0%7D%20sin%28%5Ctheta%20-%20%20%5Calpha%20%29%20%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%5D%20-%20X%20sin%20%5Calpha%20%20%3D%200)
Hence obtain
![aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta - \alpha )}]^{2} \\ b = cos \alpha \, tan(\theta - \alpha ) + sin \alpha](https://tex.z-dn.net/?f=aX%5E%7B2%7D-bX%3D0%20%5C%5C%20where%20%5C%5C%20a%3D4.9%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%5D%5E%7B2%7D%20%5C%5C%20%20b%20%3D%20cos%20%5Calpha%20%5C%2C%20%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%20%2B%20sin%20%5Calpha%20)
The non-triial solution for X is
Answer:
![X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}}](https://tex.z-dn.net/?f=X%3D%20%5Cfrac%7Bsin%20%5Calpha%20%20%2B%20cos%20%5Calpha%20%20%5C%2C%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%7B4.9%20%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20%5C%2C%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%20%20%5D%5E%7B2%7D%7D%20)
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer: X = 101.5 m
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
Hence obtain
The non-triial solution for X is
Answer:
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer: X = 101.5 m
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + ( -
)² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL