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saul85 [17]
3 years ago
7

Please help me with my sample work

Mathematics
1 answer:
eduard3 years ago
8 0
33)
(5x²+3)+(15x²+2)
5x^2+15x^2+3+2
20x^2+5
5(4x^2+1)

39)
(14h^4+3h^3)-(9h^4+2h^3)
14h^4-9h^4+3h^3-2h^3
5h^4+h^3
h^3(5h+1)


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4 0
3 years ago
A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling
lara31 [8.8K]

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

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The Normal force

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N = 2773 lbs

Then equal force in magnitude and in opposite direction will car exets on the driveway

3 0
3 years ago
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