Answer:
m1=914.9kg
m2=604.9kg
m3=864.75kg
Explanation
I think we are suppose to find the mass of the crate.
The effective force that moves the body in positive x direction is 3615N
ΣFx = Σma
Then Fx=3615N
Then the masses be m1, m2 and m3
Then,
ΣF = Σ(ma)
3615=(m1+m2+m3)a
Given that a=1.516
The masses are
m1+m2+m3=, 2384.56. Equation 1
Between mass 1 and mass 2 is, F12=1387.
The effective force that pull mass 1 is 1387.
F12=m1 ×a
Therefore,
m1=F12/a
m1=1387/1.516
m1=914.9kg.
The effective force that pulls crate 1 and crate 2 is F23
F23=(m1+m2)a
Therefore
2304=(m1+m2)a
Therefore, since a=1.516
m1+m2=2304/1.516
m1+m2=1519.8kg
Since m1=914.9kg
So, m2=1519.8-m1
m2=1519.8-914.9
m2=604.9kg
Also from equation 1
m1+m2+m3=2384.56
Since m1=914.9kg and m2=604.9kg
Then, m3=2384.56-604.9-914.9
m3=864.75kg
Which type of mirror always has an image that is smaller than the object?
convex
Explanation :
It is given that, q and 4q are placed at a distance of l.
Let x is the distance where third charge is placed so that the entire three charge system is in static equilibrium.
Equilibrium means the net force acting on the system of charges is equal to zero. Let Q is the third charge.
So, 
On solving,

For magnitude :

using 

So, a charge of -4q/9 is at a distance of l/3 is placed. It is placed to the right of +q.
Explanation:
Formula to determine the critical crack is as follows.

= 1,
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and, 
= 427.5
Hence, we will calculate the critical crack length as follows.
a = 
= 
= 
Therefore, largest size is as follows.
Largest size = 2a
= 
= 
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is
.