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3 years ago

Which most likely indicate a chemical change has occurred ?

1 answer:

5 0

If a new substance suddenly appears that wasn't there originally,
then a chemical change has occurred.

Like for example, (this is the only example I can think of right now):

-- You leave your bicycle outside in the rain, and it gets wet, and
a few days later there's some rust on it.

-- You scrape off some of the rust, take it to school, give it to the
Chemistry teacher, and ask her to analyze it and tell you what it
is. Later that day, she tells you it's a substance called "Iron oxide".

-- Where did that come from ? There was no iron oxide there.
There was only iron in the bicycle, and air, and water.

-- The iron oxide formed from a chemical change when the iron
on the surface of the bike combined with some of the oxygen in
the air, and molecules of a new substance were created. (For
some reason, the presence of water makes this chemical reaction
go faster.)

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An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

SCORPION-xisa [38]

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

5 0

3 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago

According to Newton, there were two things needed for an object to fall around or orbit the earth. Label the diagram below with

iVinArrow [24]

Centripetal force and centrifugal force

4 0

3 years ago

In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug

Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0

3 years ago

An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance f

EleoNora [17]

Answer:

364days

Explanation:

Pls see attached file

Explanation:

8 0

3 years ago