An Inherited Behavior is taught to the animal after the animal is born. True False

A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea

Savatey [412]

Answer:

$41.81^{\circ}$

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector $\overrightarrow{OA}$ is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector $\overrightarrow {OA}$ and head of the vector $\overrightarrow{AB}$ must lie on the north-south line.

Now, for this condition, from the triangle OAB

$|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|$

$\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3$

$\Rightarrow \theta=\sin^{-1}\frac23$

$\Rightarrow \theta=41.81^{\circ}$

Hence, the kayaker must paddle in the direction of $41.81^{\circ}$ in the north of east direction.

3 0

3 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, radial ( $a_{r}$ ) and tangential ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

2 years ago

Describe the valence electrons in nitrogen and how it could bond to other atoms?

Gekata [30.6K]

Answer:

Each nitrogen molecule consists of two atoms of nitrogen that are bonded by a triple covalent bond. This is because each nitrogen atom has 5 valence electrons. Each atom can complete its octet by sharing three electrons.

I hope this helps!

4 0

3 years ago

Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o

Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

$F_{23} = \frac{k*q_{2}*q_{3}}{x^2}$ (Electric force of q₂ over q₃)