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m_a_m_a [10]
2 years ago
14

Someone help me please

SAT
1 answer:
mart [117]2 years ago
6 0

Answer:

download Grauthmath it will help u

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Answer:

I mean rubber protects you

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3 years ago
A(t)=(t−k)(t−3)(t−6)(t+3) Given that a(2)=0 , what is the absolute value of the product of the zeros of a?
Ugo [173]
So using a(2)=0 we can first solve for k by substituting t for 2
0 = (2-k)(2-3)(2-6)(2+3)
0 = (2-k)(-1)(-4)(5)
0 = (2-k)20
0 = 40 - 20k
-40 = -20k
k = 2

The next step would be to find all the 0s of a.
0 = (t-2)(t-3)(t-6)(t+3)
T = 2,3,6,-3

Then we find the product
2x3x6x-3 = -108

Since the problem asks for the absolute value, the answer is positive 108
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Read 2 more answers
Friction: a 50-kg box is resting on a horizontal floor. A force of 250 n directed at an angle of 30. 0° below the horizontal is
sergejj [24]

The net forces on the box parallel and perpendicular to the surface, respectively, are

∑ F[para] = (250 N) cos(-30.0°) - F[friction] = (50 kg) a

and

∑ F[perp] = F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

where a is the acceleration of the box. (We ultimately don't care what this acceleration is, though.)

To decide whether the friction here is static or kinetic, consider that when the box is at rest, the net force perpendicular to the floor is

∑ F[perp] = F[normal] - F[weight] = 0

so that, while at rest,

F[normal] = (50 kg) g = 490 N

Then with µ[s] = 0.40, the maximum magnitude of static friction would be

F[s. friction] = 0.40 (490 N) = 196 N

so that the box will begin to slide if it's pushed by a force larger than this.

The horizontal component of our pushing force is

(250 N) cos(-30.0°) ≈ 217 N

so the box will move in our case, and we will have kinetic friction with µ[k] = 0.30.

Solve the ∑ F[perp] = 0 equation for F[normal] :

F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

F[normal] - 125 N - 490 N = 0

F[normal] = 615 N

Then the kinetic friction felt by the box has magnitude

F[k. friction] = 0.30 (615 N) = 184.5 N ≈ 185 N

8 0
2 years ago
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