Ask question

Ask question

All categories

3 years ago

15

Jamal combined baking soda and vinegar but he is having trouble determining if what he observed is a chemical reaction or a phys

ical change. a.change in color
b. absorption of heat
c. no change on properties
d. formation of new substances

2 answers:

maria [59]3 years ago

4 0

D is the answer............

KonstantinChe [14]3 years ago

4 0

D.<span>formation of new substances i hope this helps you

</span>

You might be interested in

A 600 N astronaut travels to an asteroid where the gravitational force is one-hundredth that of Earth. What is the astronaut's w

Ganezh [65]

Answer:

A) 6N

Explanation:

the weight of the astronaut on earth can be calculated with the formula

Weight = mass * gravity of earth

Since the gravitational force is one-hundredth of Earth, the formula should be

Weight = mass * (gravity of earth / 100)

Weight = (mass * gravity of earth) / 100

Since you already know the weight, you only need to divide by 100

Weight = (mass * gravity of earth) / 100

Weight = 600N / 100

Weight = 6N

4 0

3 years ago

What would the final volume of 40 L of gas at 80 pascals be if the pressure increases to 130 pascals?

melomori [17]

Answer:

Final volume, V2 = 24.62 L

Explanation:

Given the following data;

Initial volume = 40 L

Initial pressure = 80 Pa

Final pressure = 130 Pa

To find the final volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the equation, we have;

$80 * 40 = 130V_{2}$

$3200 = 130V_{2}$

$V_{2} = \frac {3200}{130}$

$V_{2} = 24.62$

Final volume, V2 = 24.62 Liters

5 0

3 years ago

PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!

SVETLANKA909090 [29]

<u>Answer:</u>

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 $m/s^2$

<u>Explanation:</u>

1) Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2) Average speed = Total distance covered / Time taken for that distance to cover.

Total distance covered = 2+0.5+2.5 = 5 km

Time taken = 3 hours

Average speed = 5/3 = 1.66 km /hour

3) Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

Change in velocity = 14 - 6 = 8 m/s

Time = 4 seconds

So acceleration = 8 / 4 = 2 $m/s^2$

6 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0

3 years ago

In Ch. 1.6, the authors point out that interstellar space is not actually as empty as it seems. There is actually a lot of diffu

wel

Answer:

very small solid particles called interstellar dust.

Explanation:

In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.

Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).

Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.

Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.

6 0

3 years ago