Answer:
The appropriate response is "
". A further explanation is described below.
Explanation:
The torque (
) produced by the force on the dam will be:
⇒ 
On applying integration both sides, we get
⇒ 
⇒ 
⇒ ![=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]](https://tex.z-dn.net/?f=%3DpgL%5B%5Cfrac%7Bh%5E3%7D%7B2%7D%20-%5Cfrac%7Bh%5E3%7D%7B3%7D%20%5D)
⇒ 
D,f,g,h,i,a,e,c,j. I’m sure that it
Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.
Answer:
d = (75 i ^ + 93 j ^ + 27 k ^) m
, d2 = (900 i ^ + 1116 j ^ + 324 k ^) m
Explanation:
The two objects are in circular orbit together, therefore with the same angular velocity, after the launch they move with the relative velocity, so we can use the kinematic relation
v = d / t
d = v t
Reduce time to units SI
t = 5 min (60 s / 1 min) = 300 s
X axis
x = vₓ t
x = 0.25 300
x = 75 m
Y axis
y =
t
y = 0.31 300
y = 93 m
Z axis
z=
t
z = 0.09 300
z = 27 m
d = (75 i ^ + 93 j ^ + 27 k ^) m
For the time of 1 h
t2 = 1 h (3600s / 1 h) = 3600
x2 = 900 m
y2 = 1116 m
z2 = 324 m
d2 = (900 i ^ + 1116 j ^ + 324 k ^) m