The primary purpose(s) of an antenna is to: . Select all that apply.

Encode a wave with the information to be sent.
Decode a wave that has been received and turning it into a useful form.
Receive a wave from the air or other surroundings.
Send a wave into the air or other surroundings.

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Ответ:

alexcuevaz90

25.09.2020

Physics

Free body diagram: force of gravity $f_g$ pointing down, with normal force $f_n$ with same magnitude, pointing up.in horizontal direction, there is only friction force $f_f$ , pointing opposite to direction of motion.so the net force comes from friction force (gravity and normal cancel out).=== $f_g = mg = (20\text{ kg})(9.81\text{ m/s}^2) = 196.2 \text{ n}$ normal force is $f_n = -f_g = -196.2 \text{ n}$ friction force is, with $\mu = 0.3$ $\begin{aligned} f_f & = \mu f_n \\ & = 0.3 \cdot (-196.2 \text{ n})\\ & = -58.86\text{ n} \end{aligned}$ therefore, $f_{net} = -58.86\text{ n}$ and $\begin{aligned} f_{net} & = ma \\ a & = \frac{f_{net}}{m} \\ a & = \frac{-58.86\text{ n}}{20\text{ kg}} \\ a & = -2.943\text{ m/s}^2 \\ \end{aligned}$ we have kinematic equation $\begin{aligned} v^2 & = v_0^2 + 2ad\\ v^2- v_0^2 & = 2ad \\ d & = \dfrac{v^2- v_0^2}{2a} \end{aligned}$ if initial velocity $v_0 = 10\text{ m/s}$ final velocity $v = 0\text{ m/s}$ acceleration $a & = -2.943\text{ m/s}^2$ then $\begin{aligned}d & = \dfrac{v^2- v_0^2}{2a}\\ & =\dfrac{0^2- 10^2}{2(-2.943)} \\ & = \text{$16.989$ m} \end{aligned}$ 16.989 m, or 20 m for one significant figure.

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