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:
Here,
x + 25° + 3x + 95° + 80°=360° (Sum of angles of a quadrilateral is 360°)
or,x+3x + 25° + 95° + 80° =360°
or,4x + 200 = 360°
or,4x = 360 - 200
or,4x = 160
or,x = 160÷4
or,x = 40
Now,
Angle K = (x + 25)° = 40 + 25° = 65°
Angle L = 3x° = 3 × 40° = 120°
14 hundreds, 10 tens, and 2 ones.
