What type of stimulus do we tend to be particularly aware of ?

If a bullet travels at 593.0 m/s, what is its speed in miles per hour?

Ksenya-84 [330]

We have $1 \; mile = 1609.34\; meters$ . So, $1 \; meter = \frac{1}{1609.34} \;mile$ .

$1 \; hour = 3600 \; s$ . So $1 \; s = \frac{1}{3600} \; hour$ .

Thus we can convert the units of the given quantity.

That is,

$593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\
593\;m/s=1,326.51\;miles/hour$ .

The quantity is converted to the required units.

7 0

3 years ago

Lunar missions have revealed that the moon has:

myrzilka [38]

That the moon has soil within its shadowy craters rich and useful material

5 0

3 years ago

Cual es el deporte que le da fortaleza y flexibilidad al cuerpo

AlexFokin [52]

Answer:

Aesthetic sports

Explanation:

Aesthetic sports are the one's that need well-developed physical qualities such as strength, agility, stamina, flexibility, and technical knowledge and artistry, in addition to technical ability and artistry. Elite athletes in these sports generally have a low abdominal fat , and the ranking is subjective.

In aesthetic sports like gymnastics, swimming, and figure skaters, dynamic and proactive flexibility is required.

4 0

3 years ago

What should stainless steel be considered when using it?

disa [49]

Strength, ductility, and toughness are the three top mechanical qualities that should be given importance. Stainless steel comes with 10–30% chromium as its alloying element. It is this element that helps it resist corrosion.

4 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$