The simplest type of vacuum tube is called a

Natali5045456 [20]

it was too much water inside the clouds and the tornado happened

6 0

3 years ago

I was in Anchorage Alaska during the Summer Solstice. The sun rose around 3am and set around 1 am the next day-- almost 24 hours

m_a_m_a [10]

Answer:

Daylight hours would be shorter.

Explanation:

If there were no tilt of the axis at all, every place on the planet except the north and south poles would have 12 hours of daylight and 12 hours of night every day of the year.

At a 10° tilt, the arctic circle and antarctic circles would be a little less than half the distance from the poles as they are today.

3 0

3 years ago

What sphere on earth includes mountains

Lostsunrise [7]

I believe it is lithosphere

3 0

3 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

3 years ago

Determine the linear velocity of an object with an angular velocity of 5.9 radians per second at a distance of 12 centimeters fr

Andre45 [30]

The linear velocity of a rotating object is the product of the angular velocity and the radius of the circular motion. Angular velocity is the rate of the change of angular displacement of a body that is in a circular motion. It is a vector quantity so it consists of a magnitude and direction. From the problem, the angular velocity is 5.9 rad per second and the radius is given as 12 centimeters. We calculate as follows:

Linear velocity = angular velocity (radius)
Linear velocity = 5.9 (12 ) = 70.8 cm / s

The linear velocity of the body in motion is 70.8 centimeters per second or 0.708 meters per second.

7 0

3 years ago