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3 years ago

A change in color during a reaction is often a sign that a chemical change has occurred.

Physics

2 answers:

Inga [223]3 years ago

5 0

<h3><u>Answer;</u></h3>

The above statement is <u>True</u>;

A change in color during a reaction is often a sign that a chemical change has occurred.

<h3><u>Explanation</u>;</h3>

A change in color indicates the occurrence of a chemical change. This means a chemical reaction is occurring. The rearrangement of electrons between substances results in new substances being formed.
Additionally, changes in color, formation of a precipitate, temperature change or a change in pH also indicate the occurrence of a chemical reaction.

guajiro [1.7K]3 years ago

3 0

True, also fizzing and heat being released

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Sedbober [7]

Answer:

Explanation:

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3 years ago

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w

amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =\frac{d}{t} (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

⇒ t = 0.2 ms (with two significative figures)

6 0

2 years ago

an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a s

valkas [14]

Answer:k=28.29 kN/m

Explanation:

Given

mass $m =7700 kg$

height from which Elevator falls $h=32 m$

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

$\frac{kx^2}{2}=mg(h+x)$ ----------1

also maximum acceleration is 5g

thus

$mg-kx=ma$

here $a=-5g$

$kx=mg-m(-5g)=6mg$

$x=\frac{6mg}{k}$

Substitute x in equation 1

$0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})$

$18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}$

$k=12\cdot \frac{mg}{h}$

$k=12\times \frac{7700\times 9.8}{32}$

$k=28.29 kN/m$

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3 years ago

If two negative charges decrease 1/2 times it's original charge then the force will become how many times stronger

vovangra [49]

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3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago