F = 52000 N
m = 1060 kg
a= F/m = 52000 N/1060 kg = 49.0566 m/s^2
 
        
             
        
        
        
4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.
        
             
        
        
        
Among the choices above, the one
that is most closely related to an activated complex is the transition state. The
answer is letter D. This formation forms quickly and does not stay in a way
compound is. It usually forms during the enzyme – substrate reaction.
 
        
             
        
        
        
See the graph in attachment
Explanation:
In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.
In the graph, we see that the initial velocity at time t = 0 is

and it is negative, so below the x-axis.
Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).
After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.
Learn more about velocity:
brainly.com/question/5248528
#LearnwithBrainly
 
        
             
        
        
        
Answer: 
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m 
and f =  v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69 
given that A = 2.20 mm = 2.2×10⁻³
so  = A × ω
 = A × ω
 = 2.2×10⁻³ × 2722.69 m/s
 = 2.2×10⁻³ × 2722.69 m/s 
 =  5.9899 m/s
 =  5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin(  0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
 = A' × ω
 = A' × ω
 = 1.555×10⁻³ × 2722.69
 = 1.555×10⁻³ × 2722.69 
 = 4.2338 m/s
 = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s