All of the following are regulated by the medulla except _________

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago

From the center of the Earth to the moon, what should the orbital radius of such satellite be in order to stay over the same poi

yulyashka [42]

In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.

But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.

The Moon has nothing to do with any of this.

3 0

3 years ago

Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl

ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
Both movements are independent each other, due to they are perpendicular.
In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

$v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)$

Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

$t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)$

In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)$

In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

$v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)$

Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

$\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)$

Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.

4 0

3 years ago

In the Bohr model of the hydrogen atom, an electron orbits a proton in a circular orbit of radius 0.53 X 10^-10 m. What is elect

maxonik [38]

Answer:

a) 27.2 V

b)27.2 V

Explanation:

Charge of the electron =charge of the proton = q = 1.6 × 10⁻¹⁹ C

Radius = r = 0.53×10⁻¹⁰ m

Electric Potential = V = k q/r

k = 9 ×10⁹ N m²/C² = Coulomb's constant.

V = (9 ×10⁹)(1.6 × 10⁻¹⁹)/( 0.53×10⁻¹⁰) = 27.2 V

b) Potential Energy of the electron = k q × q / r

= [(9 ×10⁹)(1.6 × 10⁻¹⁹)(1.6 × 10⁻¹⁹) / (0.53×10⁻¹⁰)] / (1.6 × 10⁻¹⁹) eV,

since 1 electron volt = (1.6 × 10⁻¹⁹)joules

= 27.2 eV

7 0

4 years ago

Three identical satellites, X, Y, and Z are in outer space. The distance from satellite X to Y is 450 km, the distance from Y to

hodyreva [135]

Gravity is stronger the closer you get. It is D.

3 0

3 years ago

All of the following are regulated by the medulla except __________.