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frez [133]
2 years ago
15

A student pulls a block over a rough surface with a constant force fp that is at an angle θ above the horizontal, as shown above

. If fp remains constant but the angle θ is increased, which of the following is true at some later time?.
SAT
1 answer:
Lesechka [4]2 years ago
6 0
The force of friction between the block and surface will decrease.
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37 cos(x2) dx 0 Do the following. (a) Find the approximations T8 and M8 for the given integral. (Round your answer to six decima
dusya [7]

The approximations T8 and M8 for the given integral are:

  • T8 = 33.386321; and
  • M8 = 33.50794

<h3>What is an integral?</h3>

An Integral is a variable of which a given function is the derivative, i.e. it gives that function when differentiated and may express the area under the curve of the function's graph.

<h3>What is the explanation to above answer?</h3>

Given:

F(x) = 37 cox (x²)

Internal = [0,1] n = 8 in Δ x = 1/8

The sub intervals are:

[0, 1/8], [1/8, 2/8], [2/8, 3/8], [ 3/8, 4/8], [ 4/8, 5/8], [ 5/8, 6/8], [6/8, 7/8], [ 7/8, 1]

The mid points are given as:

1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16

and X₀ = 0, X₁ = 1/8, X₂ = 2/8

Using the Trapezium Rule which states that:

\int\limits^1_0 cos(x)^{2} } \, dx = Δx/2 [f(xo) + 2f(x1) 2f(x2) + ....+ 2f(x7) + f(x8)]

= 1/1Q[f(0) + 2f (1/8) + 2f(2/8) + ....+ 2f(7/8) + f(1)]

= 0.902333

Now

T8 = \int\limits^1_0 {37Cos(x)^{2} } \, dx

= 37\int\limits^1_0 {(0.902333)} } \, dx

= 37 (0.902333)

T8 = 33.386321

It is to be noted that the midpoints rule is given as;

\int\limits^1_0 {Cos(x)^{2} } \, dx  = Δx [f(1/16) + (3/16) + .... + f(15/16)]

= 1/8[f(1/16) + f (3/16) + f(5/16) + f(7/16) + f(9/16) + f(11/16) + f(13/16) + f(15/16)]

= 0.905620

From the above,

M8 = \int\limits^1_0 {37 Cos(x)^{2} } \, dx

= 37\int\limits^1_0 {Cos(x)^{2} } \, dx

= 37 (0.905620)

M8 = 33.50794

Learn more about integral at;
brainly.com/question/19053586
#SPJ1

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