The two possible angles obtained by using the qudratic equation are;
θ
= 15.10° and θ2 = 73.51°
Given, speed of water =
= 50ft/s
For the motion along x direction, time period can be calculated as follows:

35 = (50 × cosθ) t
t = 0.64 / cosθ
For the motion in y direction, an equation can be obtained as follows:


θ) 
Plugging in the values we get:

θ) 
-20 = -32tanθ - 10.304
θ
Upon solving the above quadratic equation, we get,
tanθ = 0.27 , -3.38
Therefore,
tanθ
= 0.27
θ
= 15.10°
and, tanθ
= -3.38
θ
= 73.51
Learn more about quadratic equation here:
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Gurlll same I hate physics
Answer:
900 cm/s or 9 m/s.
Explanation:
Data obtained from the question include the following:
Length (L) = 30 cm
frequency (f) = 60 Hz
Velocity (v) =.?
Next, we shall determine the wavelength (λ).
This is illustrated below:
Since the wave have 4 node, the wavelength of the wave will be:
λ = 2L/4
Length (L) = 30 cm
wavelength (λ) =.?
λ = 2L/4
λ = 2×30/4
λ = 60/4
λ = 15 cm
Therefore, the wavelength (λ) is 15 cm
Now, we can obtain the speed of the wave as follow:
wavelength (λ) = 15 cm
frequency (f) = 60 Hz
Velocity (v) =.?
v = λf
v = 15 × 60
v = 900 cm/s
Thus, converting 900 cm/s to m/s
We have:
100 cm/s = 1 m/s
900 cm/s = 900/100 = 9 m/s
Therefore, the speed of the wave is 900 cm/s or 9 m/s.
A: Buoyant force is equal to the weight
Answer:

Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,

- separation distance of capacitor 1,

- quantity of charge on capacitor 2,

- quantity of charge on capacitor 1,

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air

From eq. (1)
For capacitor 2:

For capacitor 1:

![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:


& for capacitor 1:


![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)
