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leonid [27]
4 years ago
10

The function L = 0.8T2 models the relationship between L, the length in feet of a pendulum, and T, the period in seconds of the

pendulum. Which value is closest to the period in seconds for a pendulum that is 30 ft long?
A) 5.4 s
B) 4.9 s
C) 6.8 s
D) 6.1 s
Please help A.S.A.P
Mathematics
1 answer:
gizmo_the_mogwai [7]4 years ago
7 0
<span>L = (0.8) T^2
 
T^2 = L / 0.8
 
T = sqrt{ L / 0.8 }
 
T = sqrt{ 30 / 0.8 }
 
T = sqrt{ 37.5 }
 
T = 6.1 seconds
 
The correct answer is D) 6.1 seconds</span>
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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track
Aloiza [94]

Answer:

Approximately 0.31\; \rm m, assuming that g = 9.81\; \rm N \cdot kg^{-1}.

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned} .

Weight of the slide:

\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}.

Normal force between the sled and the slope:

\begin{aligned}F_{\rm N} &= W\cdot  \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}.

Calculate the kinetic friction between the sled and the slope:

\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}.

Assume that the sled and its passenger has reached a height of h meters relative to the base of the slope.

Gain in gravitational potential energy:

\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}.

Distance travelled along the slope:

\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}.

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}.

In other words, the sled and its passenger would have lost (approximately) ((137 + 68.1)\, h)\; {\rm J} of energy when it is at a height of h\; {\rm m}.

The initial amount of energy that the sled and its passenger possessed was \text{KE} = 63\; {\rm J}. All that much energy would have been converted when the sled is at its maximum height. Therefore, when h\; {\rm m} is the maximum height of the sled, the following equation would hold.

((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}.

Solve for h:

(137 + 68.1)\, h = 63.

\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}.

Therefore, the maximum height that this sled would reach would be approximately 0.31\; \rm m.

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Answer:

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Step-by-step explanation:

multiple ways to prove this.

e.g. since the angle between the two lines from the center of the circle to the 2 tangent touching points is 90 degrees (that is the meaning of these 90 degrees here as the angle of the circle segment defined by the 2 tangent touching points and the circle center), the tangents have the same "behavior" as tan and cot = the tangents at the norm circle at 0 and 90 degrees. they hit each other outside of the circle again at 90 degrees.

another way

imagine the two right triangles of the tangents crossing point to the circle center and the tangent/circle touching points.

the Hypotenuse of each triangle is cutting the 90 degree angle of the circle segment exactly in half (due to the symmetry principle). so the angle between radius side and Hypotenuse is 90/2 = 45 degrees.

that means also the angle of such a right triangle at the tangent crossing point is 45 degrees (as the sum of all angles in a triangle must be 180, we have the remainder of 180 - 90 - 45 = 45 degrees).

the angles of both right triangles at that point are the same, and so we can add 45+45 = 90 degrees for the total angle at the tangent crossing point.

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Step-by-step explanation:

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