Answer:
4
Explanation:
We are given that
K.E at x=0 m=20 J
K.E at x=3 m=11 J
We have to find the value of c.
By work energy theorem
Work done=Change in kinetic energy
W=
![W=[\frac{cx^2}{2}-x^3]^{3}_{0}](https://tex.z-dn.net/?f=W%3D%5B%5Cfrac%7Bcx%5E2%7D%7B2%7D-x%5E3%5D%5E%7B3%7D_%7B0%7D)
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
Answer:
Explanation:
The equivalent of Newton's second law for rotational motions is:
where
is the net torque applied to the object
I is the moment of inertia
is the angular acceleration
In this problem we have:
(net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
is the moment of inertia
Solving for , we find the angular acceleration:
Answer:
The magnitude of Force is 8.58×10⁵N and direction is upwards
Explanation:
The work beam does on the pile driver is given by
W=(FCos180°)Δx= -F(0.088m)
From work energy theorem
Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m
So
The magnitude of Force is 8.58×10⁵N and direction is upwards
