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3 years ago

What are three physical properties of gases

1 answer:

7 0

Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.

Explanation:

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I WILL GET BRAINLIEST PLS HELP A student attempts to push a box across the floor. The student decides to create a free-body diag

MissTica

Answer: The gravitational

Explanation: The student is pushing the box so u have to have gravitational force so it could move

5 0

2 years ago

How do the chemical properties of the halogens compare to those of the noble gases?

serg [7]

Halogens<span> are extremely reactive elements because they need one more electron to gain a full octet of valence electrons, whereas the </span>noble gases<span>are extremely unstable because they already have their full octet.</span>

8 0

3 years ago

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

lbvjy [14]

Answer:

The beam of light is moving at the peed of:

$\frac{dy}{dt} = \frac{80\pi}{3}$ km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, $\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi$ (1)

Now, in the right angle in the given fig.:

$tan\theta' = \frac{y}{3}$

Now, differentiating both the sides w.r.t t:

$\frac{dtan\theta'}{dt} = \frac{dy}{3dt}$

Applying chain rule:

$\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}$

$sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}$

Now, using $tan\theta = \frac{1}{m}$ and y = 1 in the above eqn, we get:

$(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}$

Also, using eqn (1),

$8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}$

$\frac{dy}{dt} = \frac{80\pi}{3}$

7 0

2 years ago

ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo

vaieri [72.5K]

Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement = 2m + (-3)m. Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement is 1m behind his original starting point.

4 0

2 years ago

A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi

Ahat [919]

<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = ?

Time, t = 6.8 s

Displacement, s = 1/4 mi = 400 meters

Substituting

s = ut + 0.5 at²

400 = 0 x 6.8 + 0.5 x a x 6.8²

a = 17.30 m/s²

Now we have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = ?

Time, t = 6.8 s

Acceleration, a = 17.30 m/s²

Substituting

v = u + at

v = 0 + 17.30 x 6.8

v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0

3 years ago