Answer:
Approximately 
(assuming that 
.)
Explanation:
Let 
denote the force that this spring exerts on the object. Let 
denote the displacement of this spring from the equilibrium position.
By Hooke's Law, the spring constant 
of this spring would ensure that 
.
Note that the mass of the object attached to this spring is 
. Thus, the weight of this object would be 
.
Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, 
.
The spring in this question was stretched downward from its equilibrium by:
.
(Note that is negative since this displacement points downwards.)
Rearrange Hooke's Law to find in terms of and :
.
Answer:
F1=26N and F2=09N ..this is from the two simultaneously equations
11m/s Bc of the fact that he sees her running at 11m/s
Answer:
Acceleration of Sea Lion is 4.41 g
This is 49% of maximum jet acceleration given as a = 9g
Explanation:
As we know that the radius of the circular loop is given as
R = 0.37 m
The speed of the fish is given as
Now the centripetal acceleration of the sea lion is given as
as we know that
so we have
Now Percentage of this acceleration wrt maximum jet acceleration is given as
%
Answer:
181.54 K
Explanation:
From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then
T2=v2t1/ v1
Given information
V1 435 ml
V2 265 ml
T1 298K
Substituting the given values then
T2=265*298/435=181.54 K
