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2 years ago

Why do noble gasses rarely react with other elements?

1 answer:

3 0

Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.

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Describe the stratosphere.

Leviafan [203]

1. The stratosphere is above the troposphere. This layer of the atmosphere is where planes fly. At the top of the stratosphere, there is a ozone layer.

2. The mesosphere is above the stratosphere. Temperatures drastically drop in the mesosphere. It is the middle layer of the atmosphere.

3. Here are the layers of the atmosphere:

Troposphere
Stratosphere
Mesosphere
Thermosphere
Exosphere

Hope this helps you!

8 0

3 years ago

Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

$T_e=2\pi \sqrt{\frac{L}{g_e}}$

where

$g_e = 9.8 m/s^2$ is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

$T_m=2\pi \sqrt{\frac{L}{g_m}}$

where

$g_m = 1.6 m/s^2$ is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

$\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}$

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0

3 years ago

A ball is thrown horizontally from the top of

sesenic [268]

Answer:

42.5 m/s

Explanation:

Given:

x₀ = 0 m

x = 62 m

y₀ = 80 m

y = 0 m

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: v

First, find the time it takes to land.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (80 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 4.04 s

Find the horizontal component vₓ:

x = x₀ + vₓ t − ½ aₓ t²

(62 m) = (0 m) + vₓ (4.04 s) − ½ (0 m/s²) (4.04 s)²

vₓ = 15.3 m/s

Find the vertical component vᵧ:

vᵧ = aᵧ t + v₀ᵧ

vᵧ = (-9.8 m/s²) (4.04 s) + (0 m/s)

vᵧ = -39.6 m/s

Find the speed using Pythagorean theorem:

v = √(vₓ² + vᵧ²)

v = √((15.3 m/s)² + (-39.6)²)

v = 42.5 m/s

3 0

3 years ago

Which poison was used in the Jonestown massacre?

olga nikolaevna [1]

That would be Cyanide.

Hope this helps! (:

3 0

4 years ago

Read 2 more answers

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago