Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
 
        
             
        
        
        
Answer:
B
Explanation:
OOf we are doing this stuff atm
So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B
 
        
                    
             
        
        
        
Acid it is i believe........
        
             
        
        
        
Answer:
The x represents the reference point on a motion map
Explanation:
-Motion maps are another way to represent the motion of an object. (other representations are graphical and mathematical models)
 
        
                    
             
        
        
        
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)