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3 years ago

10

If the equation on the board had shown 3 atoms of carbon on the reactants side, how many atoms of carbon would need to be repres

ented on the products side? 2 3 4

2 answers:

Sholpan [36]3 years ago

6 0

3

Explanation:

If the equation on the board had shown 3 atoms of carbon on the reactants side, it would need 3 atoms of carbon on the product side to balance it.

This is in accordance to the law of conservation of mass which states that "matter is neither created nor destroyed during the course of chemical reaction, they are only transformed from one form to another".

We expect the same number of atoms on both sides of the equation. For example:

Fe + S → FeS

1 iron atom combines with 1 sulfur atom to produce FeS which contains 1 iron and sulfur atoms.

Learn more:

Chemical equation brainly.com/question/2924195

#learnwithBrainly

EastWind [94]3 years ago

5 0

Answer:

3

Explanation:

i did the test and confirmed this answer

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If I push on the wall with 75 Newtons of force, the wall will push back with __

Scilla [17]

Answer: 0

Explanation:

75 newtons will push back canceling it out make it 0

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3 years ago

If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil

AfilCa [17]

Answer:

Explanation:

Let the volume of air be V. at atmospheric pressure, that is 10⁵ Pa

At 20 m below surface pressure will be

atmospheric pressure + hdg

10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa

At this pressure volume V becomes V/ 2.96

This volume will last 1/2.96 times time that is 60/2.96 = 20.27 minutes.

8 0

3 years ago

The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe

Hoochie [10]

Answer:

579600J

Explanation:

Given parameters:

Height of the building = 828m

Weight of the man = 700N

Unknown:

Work done by the man = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

Work done = Weight x height = 700 x 828

Work done = 579600J

6 0

2 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

2 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago