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3 years ago

What are noble gases?

Physics

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

noble gases are basically a group of gases that are similar in their chemical compounds, theres six of them : helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).

~batmans wife dun dun dun.....

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The diagram below shows a wave with its wavelength indicated in red.

Shtirlitz [24]

Answer:

answer is D.it will decrease

8 0

3 years ago

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1) A plane's velocity increases from 40 m/s to 100 m/s over a 10 second interval. What is the plane's average acceleration for t

Yakvenalex [24]

Answer:

average acceleration = 6 $\frac{m}{s^2}$

Explanation:

Recall that the average acceleration $(a)$ is defined by the change in velocity from an initial velocity $(v_i)$ , to a final velocity $(v_f)$ over the time (t) it took that change to happen. Then, in mathematical terms this is:

$a=\frac{v_f-v_i}{t}$

with our information this becomes:

$a=\frac{v_f-v_i}{t} = \frac{100-40}{10}=6\,\frac{m}{s^2}$

8 0

3 years ago

Most subcultures do not reject all of the values and practices of the larger society. Please select the best answer from the cho

iren [92.7K]

<h2><u>Answer:</u></h2>

The statement is True.

<h3><u>Explanation:</u></h3>

A subculture is a gathering of individuals inside a culture that separates itself from the parent culture to which it has a place, frequently keeping up a portion of its establishing standards.

Subcultures build up their own standards and qualities with respect to social, political and sexual issues

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3 years ago

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How many newtons of force are represented by the following amount: 3 kg.m/sec2

alekssr [168]

I believe the answer is 3 Newtons!

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3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

3 years ago