no BECQUSE POSUM BROOB SHSHSJ
To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,
Here,
v = Velocity
= Linear density (Mass per unit length)
T = Tension
Rearranging to find the Period we have that
As we know that speed is equivalent to displacement in a unit of time, we will have to
Therefore the tension is 5.54N
Answer:
speed wind Vw = 54.04 km / h θ = 87.9º
Explanation:
We have a speed vector composition exercise
In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south
Let's add the vectors in each coordinate axis
X axis (East-West)
-Xvion - Xw = -119
Xw = -Xavion + 119
Xw = 119 -108
Xwi = 1 km
Calculate the speed for time of t = 0.5 h
Vwx = Xw / t
Vwx= 1 /0.5
Vwx = - 2 km / h
Y Axis (North-South)
Y plane - Yi = -27
Y plane = 0
Yw = 27 km
Vwy = 27 /0.5
Vwy = 54 km / h
Let's use the Pythagorean theorem and trigonometry to compose the answer
Vw = √ (Vwx² + Vwy²)
Vw = R 2² + 54²
Vw = 54.04 km / h
tan θ = Vwy / Vwx
tan θ = 54/2 = 27
θ = Tan⁻¹ 1 27
θ = 87.9º
The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º west from the south
I don't think the shaping of the beach or hill would be considered
weathering, especially since it says "as smaller particles are moved
away". This one is just talking about where the particles decide to
gather and make a shape.
'C' is really talking about weathering, where rocks are broken up into
stones, and the stones into smaller pieces that can be blown away
on the wind. THAT's weathering.
Answer:
See the answers below.
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity = 10 [m/s]
Vo = initial velocity = 40 [m/s]
t = time = 5 [s]
a = acceleration [m/s²]
Now replacing:
![10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]](https://tex.z-dn.net/?f=10%3D40-a%2A5%5C%5C40-10%3Da%2A5%5C%5C30%3D5%2Aa%5C%5Ca%3D6%5Bm%2Fs%5E%7B2%7D%5D)
Note: The negative sign in the above equation means that the velecity is decreasing.
2)
To solve this second part we must use the following equation of kinematics.
where:
x = distance [m]
![(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]](https://tex.z-dn.net/?f=%2810%29%5E%7B2%7D%20%3D%2840%29%5E%7B2%7D%20-2%2A6%2Ax%5C%5C100%3D1600-12%2Ax%5C%5C12%2Ax%3D1600-100%5C%5C12%2Ax%3D1500%5C%5Cx%3D125%5Bm%5D)
