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Blababa [14]
4 years ago
12

Newton's second law says that force (f) is equal to mass (m) times acceration (a). A scientist wants to calculate the force of a

n object where the acceleration of gravity (g) is 9.8 m/s^2. Use the function to calculate the force for an object with a mass (m) of 0.29 kilograms.
A) 9.51 Newton's
B)0.029 Newton's
C) 34 Newton's
D) 2.8 Newton's
Mathematics
1 answer:
Rina8888 [55]4 years ago
7 0
It would be D. 0.29 x 9.8 = 2.842
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Find the arc length and area of a sector with radius R =60 inches and central angle equals 30° round your answer to the nearest
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Answer:

(a)31.42 inches

(b)942.48 Square Inches

Step-by-step explanation:

Given a sector of a circle with the following dimensions:

Radius of the circle =60 inches

Central Angle of the sector =30°

(a)Arc Length

Arc Length =\dfrac{\theta}{360^\circ}X2\pi r

=\dfrac{30}{360^\circ}X2*60*\pi\\\\=10\pi\\\\=31.42$ Inches (correct to the nearest hundredth)

(b)Area of the sector

Area of the sector =\dfrac{\theta}{360^\circ}X\pi r^2

=\dfrac{30}{360^\circ}X\pi*60^2\\\\=300\pi\\\\=942.48$ Square Inches (correct to the nearest hundredth)

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The answer to this question is letter B
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Find the perimeter of pentagon STUVW WITH VERTICES S(0,0) T(3,-2) U(2,-5) V(-2,-5) W(-3,-2)
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Use the distance formula.

\sqrt{( x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}

 
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Points S and T
\sqrt{(3 - 0)^2 + (-2 - 0)^2}

\sqrt{(3)^2 + (-2)^2}

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\sqrt{13}

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Points T and U
\sqrt{(3 - 2)^2 + (-2 + 5)^2}

\sqrt{(1)^2 + (3)^2}

\sqrt{1+9}

\sqrt{10}

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Points U and V
\sqrt{(2+2)^2 + (-5 + 5)^2}

\sqrt{(4)^2 + (0)^2}

\sqrt{16}

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Points V and W
\sqrt{(-2+3)^2 + (-5 + 2)^2}

\sqrt{(1)^2 + (-3)^2}

\sqrt{2+9}

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Add all these together.

3.3 + 3.1 + 4 + 3.1 + 3.6
≈17
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4 years ago
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