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:
5 + 2x = 6
2x = 6-5
2x = 1
x = 0.5
n -3n = 14 -4n
-2n = 14-4n (since 4n is negative, change the sign to make it positive and then add 4n to both sides of the equation)
2n = 14
n = 7
7 (5a - 4) - 1 = 14 - 8a
35a - 28 - 1 = 14 -8a
35a -29 = 14 - 8a
43a = 43
a = 1
Let

be the length of the rectangle and

be the width. In the problem it is given that

. It is also given that the area

. Substituting in the length in terms of width, we have

. Using the zero product property,

. Solving these we get the width

. However, it doesn't make sense for the width to be negative, so the width must be

. From that we can tell the length

.