Answer:
Option D
670 Kg.m/s
Explanation:
Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)
Final momentum is also mv but v being westward direction, we take it negative
Final momentum=82*-2.5= -205 Kg.m/s
Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s
Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s
Answer:
304.86 metres
Explanation:
The x and y cordinates are 
and 
respectively
The horizontal distance travelled, 
Making t the subject, 
Since 
, we substitute t with the above and obtain
Making d the subject we obtain
d=304.8584
d=304.86m
Reduce friction because friction just makes things harder
So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2
