Answer:
a) weight of the car = 2816,1 lbs
b) 2773 lbs
Step-by-step explanation:
The equilibrium force is 490 lbs. That force keep the car at rest, then
∑ Fy = 0 and ∑Fx = 0
Forces acting on the car:
The external force 490 lbs
weight of the car uknown
Normal force
sin∠10° = 0,174
cos∠10° = 0.985
∑Fx = 0 mg*sin10°- 490 = 0 ∑Fy = 0 mg*cos10° - N = 0
mg*0,174= 490
mg = 490 / 0,174
mg = 2816,1 lbs
weight of the car = 2816,1 lbs
The Normal force
mg*cos10° - N = 0 2816,1 * 0,985 = N
N = 2773 lbs
Then equal force in magnitude and in opposite direction will car exets on the driveway
Answer: I think its 63−332+45−6
Step-by-step explanation:
Let 
be the amount of food needed for one man for one day. At the beginning, we have enough food for 400 men for 40 days, i.e.
After 10 days, the 400 men will have consumed
food, impliying that the food remaining is
If 200 more men join the garrison, there are now 600 men. Each of them requires food each day, and there are 
units of food remaining. They will last
days.
