Answer:
44.95 tonnes
Explanation:
According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces
It is given that empty weight of box = 40 tons
Let the mass of the stones to be placed be = M tonnes
Thus the combined mass of box and stones = (40+M) tonnes..........(i)
Since the box will displace water equal to it's volume V we have 

Now the weight of water displaced =
is density of water = 1000kg/
Thus weight of liquid displaced =
..................(ii)
Equating i and ii we get
40 + M = 84.95
thus Mass of stones = 44.95 tonnes
The false statement about onStep is: B. The default number of steps per second is 30.
<h3>What is an onStep?</h3>
An onStep can be defined as a computerized telescope goto controller that is designed and developed to <u>animate shapes</u> while using it on a variety of mounting systems such as forks.
<h3>The characteristics of an onStep.</h3>
In Engineering, some of the characteristics that are associated with an onStep include the following:
- The onStep function can be called without user input.
- It can be used to animate shapes without user input.
- It only runs a certain number of times.
In conclusion, the default number of steps per second for onStep isn't 30.
Read more on onStep here: brainly.com/question/25619349
C. seems like the best answer. i may be wrong so don’t quote me on that
Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
Answer:
b). Occurs at the outer surface of the shaft
Explanation:
We know from shear stress and torque relationship, we know that

where, T = torque
J = polar moment of inertia of shaft
τ = torsional shear stress
r = raduis of the shaft
Therefore from the above relation we see that

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.
When r= 0, then τ = 0
and when r = R , τ is maximum
Thus, torsional shear stress is maximum at the outer surface of the shaft.