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Arada [10]
3 years ago
10

This structure shows how atoms make up sugar. The different colors represent different types of atoms. Is sugar an element, just

a molecule, or a molecule and a compound? How do you know?
Physics
2 answers:
Minchanka [31]3 years ago
5 0
Sugar is a type of carbohydrate. It turns out there is a whole class of carbohydrates called "sugars." The kind of sugar we usually think about - table sugar - is called sucrose. Sugar is made up of carbon, hydrogen and oxygen atoms. It's the way these atoms are connected that makes each type of carbohydrate different. In each molecule of table sugar there are: 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms.
The black stuff is called burnt sugar! But seriously, this is what happens when you heat or burn things that contain carbon. It reacts with oxygen and "oxidizes" (burns). The black stuff itself is mainly carbon. So is the soot inside a chimney.
Unfortunately, the bacteria in your mouth also love to eat sucrose (more than you do). When they do, they produce that yucky stuff called plaque that rots your teeth and gums.
Romashka-Z-Leto [24]3 years ago
4 0
Sugar is a compound
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A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past t
marta [7]

Answer:

Explanation:

Given that

The window height is 2m

And the window is 7.5m from the ground

Then the total height of the window from the ground is 7.5+2=9.5m

It takes the ball 0.32sec travelled pass the window.

When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')

Now using the equation of free fall during this window travels

S=ut-½gt² against motion.

S=2, g=9.81, t=0.32sec

Then,

S=u't-½gt²

2=u'×0.32-½×9.81×0.32²

2=0.32u'-0.5023

2+0.5032=0.32u'

Then, 0.32u'=2.5032

u'=2.5032/0.32

u'=7.82m/s

This is the initial velocity as the ball got the the window

Now, let analyse from the window bottom to the ground which is a distance of 7.5m

Using the equation of free fall again

v²=u²-2gH

In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,

While u is the original initial velocity from the throw of the ball

Then,

u'²=u²-2gH

7.82²=u²-2×9.81×7.5

61.146=u²-147.15

61.146+147.15=u²

Then, u²=208.296

So, u=√208.296

u=14.43m/s

The initial velocity of the ball form the throw is 14.43m/s

6 0
3 years ago
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
Which statements correctly characterize a lunar eclipse? Select all that apply.
Zepler [3.9K]

A) it occurs when earth is between the sun and the moon

6 0
3 years ago
Oxygen at 50 lbf/in.2, 200 F is in a piston/cylinder arrangement with a volume of 4 ft3. It is now compressed in a polytropic pr
Ksju [112]

Answer:

The value of heat transfer during the process Q = - 29.49 KJ

Explanation:

Given data

P_{1} = 50 \frac{lbf}{in^{2} } = 344.7 k pa

V_{1} = 4 ft^{3} = 0.113 m^{3}

T_{1} = 200 F = 366.4 K

T_{2} = 400 F = 477.6 K

Poly tropic index n = 1.2

gas constant for oxygen = 0.26 \frac{KJ}{kg K}

From ideal gas equation

P_{1} V_{1} = m R T_{1}

Put all the values in above equation we get

⇒ 344.7 × 0.113 = m × 0.26 × 366.4

⇒ m = 0.408 kg

Heat transfer in poly tropic process is given by

Q = \frac{\gamma - n}{( \gamma - 1)( n - 1)} [ {m R (T_{1} - T_{2}  ) ]

Put all the values in above formula we get

⇒ Q = \frac{1.4 - 1.2}{( 1.4 - 1)( 1.2 - 1)} [ {m R (T_{1} - T_{2}  ) ]

⇒ Q = 2.5 × 0.408 × 0.26 × ( 366.4 - 477.6 )

⇒ Q = - 29.49 KJ

This is the value of heat transfer during the process & negative sign shows that  heat is lost during the process.

3 0
4 years ago
Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
3 years ago
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