The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height
The potential energy of a spring is given by:
V = 0.5 * k * x²
k spring constant
x compression of the spring
If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:
m * g * h = 0.5 * k * x²
m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m
Solve for k.
k = 2 * m * g * h / x² = 40.8 N/m
        
             
        
        
        
Answer:
341.46miles
Explanation:
Find the diagram attachment.
To get the displacement D, we will use the cosine rule as shown;
D² = 200²+150²-2(160)(400)cos65°
D² = 40000+22500-128000cos65°
D² = 62500+54095.14
D² = 116595.14
D = √116595.14
D= 341.46 miles
Hence the plane final displacement is 341.46miles
 
        
             
        
        
        
The number that indicates typical continental conditions (regional metamorphism) is that showing schist and gneiss rocks.
<h3>What is regional metamorphism?</h3>
Regional metamorphism occurs when rocks undergoes changes as a result of high temperatures and pressure deep within the earth's crust.
Regional metamorphic rocks are usually foliated or squashed in appearance. 
Examples of regional metamorphism rocks are schist and gneiss rocks.
Therefore, the figure that indicates typical continental conditions (regional metamorphism) is that showing schist and gneiss rocks.
Learn more about regional metamorphism at: brainly.com/question/14678538
#SPJ11
 
        
                    
             
        
        
        
Answer:
 The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
 
        
             
        
        
        
The answer is A. The kinetic energy