Please help me with these

A plane leaves the airport in Galisteo and flies 160 km at 66.0 ∘ east of north; then it changes direction to fly 260 km at 49.0

user100 [1]

<u>The question doesn't have any particular requirement, but we'll compute the displacement of the plane from its initial and final landing point in the pasture </u>

Answer:

$\displaystyle |\vec{r}|=321.464\ km$

$\displaystyle \theta =-19.395^o$

Explanation:

<u>Displacement </u>

The vector displacement $\vec r$ is a measure of the change of position of a moving object. The displacement doesn't depend on the path followed, only on the final and initial positions. Its scalar counterpart, the distance, does measure the total space traveled and considers all the changes in the direction taken by the object. To find the displacement, we must add all the particular displacements by using vectors.

The plane first flies 160 km at 66° east of north. To find the vector expression of this displacement, we must know the angle with respect to the East direction or North of East. Knowing the angle East of North is 66°, the required angle is 90°-66°=34°

The first vector is expressed as

$\displaystyle \vec{r_1}=\left \langle 160^o\ cos34^o, 160\ sin34^o \right \rangle$

$\displaystyle \vec{r_1}=\left \langle 132.646, 89.471 \right \rangle$

The second displacement is 260 km at 49° South of East. This angle is below the horizontal respect to the reference, thus we use -49°.

The second vector is expressed as:

$\displaystyle \vec{r_2}=\left \langle 260\ cos(-49^o), 260\ sin(-49^o)\right \rangle$

$\displaystyle \vec{r_2}=\left \langle 170.575,-196.224\right \rangle$

The total displacement is computed as the vectorial sum of both vectors

$\displaystyle \vec{r}=\vec{r_1}+\vec{r_2}=\left \langle 132.646+170.575\right \rangle+ \left \langle89.471-196.224\right \rangle$

$\displaystyle \vec{r}=\left \langle 303.221,-106.753\right \rangle\ km$

The magnitude of the total displacement is

$\displaystyle |\vec{r}|=\sqrt{303.221^2+(-106,753)^2}$

$\displaystyle |\vec{r}|=321.464\ km$

And the direction is

$\displaystyle tan\ \theta =\frac{-106.753}{303.221}=-0.352$

$\displaystyle \theta =-19.395^o$

6 0

3 years ago

The theory of evolution is

tresset_1 [31]

one of several competing scientific models that all seem equally successful in explaining the nature of life on Earth

Explanation:

The theory of evolution is one of the several competing scientific models that all seems to successfully explain the nature of life on earth.
A theory is usually drawn from careful observation of scientific investigation.
Theories generally provides explanation into a phenomenon that occurs in nature.
They are products of careful scientific observations and experiments.
The theory of evolution is but one of them.

Learn more:

Experiments brainly.com/question/5096428

#learnwithBrainly

4 0

3 years ago

Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is

Kay [80]

Answer:

the correct option is C

Explanation:

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

3 0

3 years ago

a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?

evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

$\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}$