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:
![s_{x} = (v_{A}) _x_{} } t](https://tex.z-dn.net/?f=s_%7Bx%7D%20%3D%20%28v_%7BA%7D%29%20_x_%7B%7D%20%7D%20t)
35 = (50 × cosθ) t
t = 0.64 / cosθ
For the motion in y direction, an equation can be obtained as follows:
![s_{y} = (v_{A})_{y} t +\frac{1}{2} (a_{y} )t^{2}](https://tex.z-dn.net/?f=s_%7By%7D%20%3D%20%28v_%7BA%7D%29_%7By%7D%20%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28a_%7By%7D%20%29t%5E%7B2%7D)
![s_{y} = (-v_{A}](https://tex.z-dn.net/?f=s_%7By%7D%20%3D%20%28-v_%7BA%7D)
θ) ![} t +\frac{1}{2} (a_{y} )t^{2}](https://tex.z-dn.net/?f=%7D%20%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28a_%7By%7D%20%29t%5E%7B2%7D)
Plugging in the values we get:
![-20 = (-50_](https://tex.z-dn.net/?f=-20%20%3D%20%28-50_)
θ) ![} t +\frac{1}{2} (-32.2} )t^{2}](https://tex.z-dn.net/?f=%7D%20%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28-32.2%7D%20%29t%5E%7B2%7D)
-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:
brainly.com/question/17177510
#SPJ4
<span>Since P = V x I, a 10% reduction of power would lead to a 10% reduction in the product of voltage and current. What is left is 90% of the original power:
.9P = .9(V x I).
If we assume that current must be the same, then we can regroup the terms on the right-hand side as follows:
.9P = (.9V) x I
In this case, voltage is also reduced by 10% (100% - 90% = 10%).</span>
Answer:
or 0.07163 T into the page
Explanation:
m = Mass of particle = 10 g
a = Acceleration due to gravity = -9.8j m/s²
v = Velocity of particle = 19i km/s
q = Charge of particle = 72 μC
B = Magnetic field
Here the magnetic and gravitational forces on the particle are applied in the opposite direction so,
![F_b=F_g](https://tex.z-dn.net/?f=F_b%3DF_g)
![F_b=qvBsin\theta\\\Rightarrow F_b=qvBsin90\\\Rightarrow F_b=72\times 10^{-6}\times 19000B](https://tex.z-dn.net/?f=F_b%3DqvBsin%5Ctheta%5C%5C%5CRightarrow%20F_b%3DqvBsin90%5C%5C%5CRightarrow%20F_b%3D72%5Ctimes%2010%5E%7B-6%7D%5Ctimes%2019000B)
![F_g=ma\\\Rightarrow F_g=0.01\times -9.8](https://tex.z-dn.net/?f=F_g%3Dma%5C%5C%5CRightarrow%20F_g%3D0.01%5Ctimes%20-9.8)
![72\times 10^{-6}\times 19000B=0.01\times -9.8\\\Rightarrow B=\frac{0.01\times -9.8}{72\times 10^{-6}\times 19000}\\\Rightarrow B=-0.07163\hat k\ T](https://tex.z-dn.net/?f=72%5Ctimes%2010%5E%7B-6%7D%5Ctimes%2019000B%3D0.01%5Ctimes%20-9.8%5C%5C%5CRightarrow%20B%3D%5Cfrac%7B0.01%5Ctimes%20-9.8%7D%7B72%5Ctimes%2010%5E%7B-6%7D%5Ctimes%2019000%7D%5C%5C%5CRightarrow%20B%3D-0.07163%5Chat%20k%5C%20T)
The magnetic field is 0.07163 T into the page
Answer:
D. power
Explanation:
kg represents mass
(m/v)² represents velocity squared
Then kg·m²/s² represents mass·velocity² = <em>kinetic energy</em> or <em>potential energy</em> or <em>work</em>.
kg·m²/s³ will be the <em>rate of doing work</em>, which is power
Force can be exerted into an object with out it moving, but if you were to move the object due to force it would be considered work. (yes, you can exert force without having the object moving)