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3 years ago

Is it true the atoms in sulphur liquid are the same as the atoms in solid sulphur?

1 answer:

3 0

Yes, it is true that the atoms in liquid sulphur are the same as the atoms
in solid sulphur.

Also, if you can heat it past 832°F, it'll boil, and you'll have sulphur vapor.
Those atoms are also the same.

In any one of its three physical phases, you can also spell it as "sulfur"
without changing the nature of its atoms.

You might be interested in

If the sun suddenly ceased to shine, how long would it take earth to become dark? you will have to look up the speed of light in

sleet_krkn [62]

We are 8 light minutes from the sun. That means two things, we see the sun as it was 8 minutes ago, and we WOULD continue to see the sun for 8 minutes after it disappeared.

3 0

2 years ago

Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line

Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required: vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0

3 years ago

I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app

Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

initial velocity of the sprinter, u = 18 km/h

final velocity of the sprinter, v = 27 km/h

time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

$initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\$

The time of motion is seconds;

$t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s$

The sprinter’s average acceleration is calculated as follows;

$a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2$

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0

2 years ago

For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th

iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU

7 0

3 years ago

I got this information for a lab but I don't know how to do the hypothesis and the conclusion please can you guys help me with i

pochemuha

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

Hope this helps.

8 0

2 years ago