Answer:
Explanation:
From, the given information: we are not given any value for the mass, the proportionality constant and the distance
Assuming that:
the mass = 5 kg and the proportionality constant = 50 kg
the distance of the mass above the ground x(t) = 1000 m
Let's recall that:
Similarly, The equation of mption:
replacing our assumed values:
where
So, when the object hits the ground when x(t) = 1000
Then from above derived equation:
By diregarding
1000 + 0.981 = 0.981 t
1000.981 = 0.981 t
t = 1000.981/0.981
t = 1020.36 sec
P=W/t=2940/10=294 W
W=F•s
W=m•g•s
W=150×9.8×2
W=2940 J
Answer: The average speed is 5.71 m/s
Explanation:
The average speed in this run will be the quotient between the total distance that the runner traveled, and the total time it took.
Distance = 400m + 400m + 800m = 1600m
Time = 80s + 70s + 130s = 280s
Then the average speed is:
S = 1600m/280s = 5.71 m/s
To replace all forces on a pipe by the equivalent force
T= 8× 50 = 400lb.in
M = 16.30 = 480lb.in
F₂ = 50lb
To calculate the polar moment of inertia of shaft is
J = π/π²×(R⁴-r⁴)
= π²/2×(0.875⁴ - 0.750⁴)
= 0.423in.³
To calculate the moment of inertia
J= 1/2(J)
=1/2 (0.423)
=0.21188in.³
To calculate shear flow
Qy= 2/3(R³-r³)
= 2/3(0.875³- 0.750³)
= 0.16536in.³
To calculate the thickness of the shaft
t= R-r = 0.875 - 0.750 = 0.125 in.
The stress due to torsions is.
Tx = TR/J = 400 × 0.875/0.42376
=825. 9psi.
The stress due to bending
Qx =My/T = 480 × 0.875/0.2118
=1982.3psi
The stress due to transverse shear
Qx = VQ/I(2t)
=50 × 0.16536/0.2118× 0.250
=156.1psi