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3 years ago

An angle with a measure of 30 degrees is called

Mathematics

2 answers:

erastovalidia [21]3 years ago

5 0

An acute angle!!! It’s an angle which is less than 90

amid [387]3 years ago

4 0

A 90º degree angle is a right angle. Anything below is an acute angle and anything more than 90º is an obtuse angle.

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Out of 32 students in a class five said they rode there bikes to school. Based on these results, how many of the 800 students ri

melamori03 [73]

YOu have to set a proportion or do this easy method...

5/32= <span>0.15625
the you multiply </span><span>0.15625 by 800 because that is what you want to find out of...

that means the answer is 125

125 ride their bike to school...

hope that helps and you give good feedback

</span>

3 0

3 years ago

given that sin theta= 1/4, 0 is less than theta but less than pi/2, what is the exact value of cos theta

lapo4ka [179]

Answer:

$\cos{\theta} = \frac{\sqrt{15}}{4}$

Step-by-step explanation:

For any angle $\theta$ , we have that:

$(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1$

Quadrant:

$0 \leq \theta \leq \frac{\pi}{2}$ means that $\theta$ is in the first quadrant. This means that both the sine and the cosine have positive values.

Find the cosine:

$(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1$

$(\frac{1}{4})^{2} + (\cos{\theta})^{2} = 1$

$\frac{1}{16} + (\cos{\theta})^{2} = 1$

$(\cos{\theta})^{2} = 1 - \frac{1}{16}$

$(\cos{\theta})^{2} = \frac{16-1}{16}$

$(\cos{\theta})^{2} = \frac{15}{16}$

$\cos{\theta} = \pm \sqrt{\frac{15}{16}}$

Since the angle is in the first quadrant, the cosine is positive.

$\cos{\theta} = \frac{\sqrt{15}}{4}$

3 0

3 years ago

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.