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:
D)
Step-by-step explanation:
I don't know if there was a picture, but d will made a triangular prism. All are possible but they will look more like a dome.
it is none since there is no relation
Im pretty sure it is the first one
Y-4=(x+8) = add4 to both sides of the equation, so y= x+8+4 then add like terms. Y=x+12 final answer.
