Answer:
<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg
 k = 0.4 
 F = 200 N 
<u>To </u><u>find </u><u>-</u><u> </u> acceleration 
<u>Solution </u><u>-</u><u> </u>
F= kMA 
200 = 0.4 * 20 * acceleration 
200 = 8 * a 
a = 8/200 
a = 0.04 m s² 
<h3>a = 0.04 m s²</h3>
 
        
             
        
        
        
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction. 
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy, 
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity) 
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
 
        
             
        
        
        
<span>So when two metals of equal mass but different heat capabilities are subjected to same heat quantity, the metal with higher heat capacity have the small temperature change. Heat supplied is determined as heat capacity of the metal times the change in temperature.</span>
        
             
        
        
        
I don't know fully, but I know S waves can't travel through liquids. Hope that helps a little!