3 years ago

Hey plz help mesuggest a situations where you might need to use ear defenders

1 answer:

4 0

You would need to ear defenders (I'm assuming these are ear protectors for use with loud sounds) while firing at a gun range. You could use it for loud construction areas.

You might be interested in

1. the most suitable cutting process to use on ferrous metals such as straight carbon steel is

dangina [55]

Answer:

plasma arc

cutting

Explanation:

It is plasma arc cutting because i think it is

6 0

3 years ago

Read 2 more answers

Where would you find an inclined plane in a fan

ivanzaharov [21]

The blades that spin around in the fan, because they are flat and produce work

8 0

3 years ago

Read 2 more answers

Many planets in our solar system (Saturn, Uranus, Neptune) have rings. If these rings are made of chunks of debris, why do they

Liono4ka [1.6K]

Generally, rings form from moons, asteroids, or comets that have disintegrated due to a collision or because they got too close to their planet (Roche Limit). ... Most of the debris, however, will not have enough energy to overcome the planet's gravity and will remain in orbit around the planet.

5 0

3 years ago

A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the

yuradex [85]

Answer:

$\frac{dy}{dt}=1.2\frac{mi}{min}$

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

$tan\theta=\frac{y}{x}\\y=xtan\theta(1)$

Differentiating (1) with respect to time, we get:

$\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}\\$

$\frac{dx}{dt}=0$ since x is a constant value. Replacing:

$\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}$

5 0

4 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. part a what is the elec

Anestetic [448]

Potential energy is energy that is found in a system, grounded on the position of objects. The Coulomb (C) is the unit of charge, and the unit of electric potential is the Volt (V), which is equivalent to (J/C) or Joule per Coulomb.So the formula for this is potential = kQ / d, plugging in the given from the questions will give us:potential = 8.99e9N·m²/C² * 1.602e-19C / 0.053e-9m = 27 V

3 0

3 years ago