Answer:
T = 0.03 Nm.
Explanation:
d = 1.5 in = 0.04 m
r = d/2 = 0.02 m
P = 56 kips = 56 x 6.89 = 386.11 MPa
σ = 42-ksi = 42 x 6.89 = 289.58 MPa
Torque = T =?
<u>Solution:</u>
σ = (P x r) / T
T = (P x r) / σ
T = (386.11 x 0.02) / 289.58
T = 0.03 Nm.
Note that
1 yd = 0.9144 m
Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m
Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) = 9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%
Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.
Rate of speed that toy is moving is irelevant.
childs force is:
Fc = 2N
Fc = Ff (Ff -friction force)
Ff = a*Q
where Q is weight of the toy and a is friction
if we express a we get
a = F/Q = 2/8 = 0.25
Answer:
a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).
b) The centre of mass is still at the starting point for both bodies.
c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.
Explanation:
Linear momentum is conserved.
(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0
5×30 + 60 × v = 0
v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).
b) At time t = 10 s,
Sally has travelled 25 m and the boot has travelled 300 m.
Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.
Centre of mass = [(60)(25) + (5)(-300)]/(60+5)
= 0 m.
The centre of mass is still at the starting point for both bodies.
c) The shore is 30 m away.
Speed = (Distance)/(time)
Time = (Distance)/(speed) = (30/2.5)
Time = 12 s
Hope this Helps!!!
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
