All categories

3 years ago

The graph represents the reaction 2H2 + 02 32H20 as it reaches

Physics

2 answers:

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

C. at point 2 the concentrations of H20...O2... & H2 are all changing

toward their equilibrium concentrations

D. at point 1 more reactants are converted to products than

products are converted to reactants

Alex3 years ago

5 0

Answer:

C and D

Explanation:

You might be interested in

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

onsider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST applies

cluponka [151]

The answer is C) A girl hangs by both hands, motionless, from a trapeze.

5 0

3 years ago

An 80.0 kg skier slides down a hill shaped as shown. Assume

umka21 [38]

The height above the ground from where the skier start is 11.5 m.

<h3>Conservation of energy</h3>

The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

$h = \frac{v^2}{2g} \\\\h = \frac{15^2}{2 \times 9.8} \\\\h = 11.5 \ m$

Thus, the height above the ground from where the skier start is 11.5 m.

Learn more about conservation of energy here: brainly.com/question/166559

8 0

2 years ago

Dmitry [639]

Answer: 176.4 J

Explanation:

5 0

3 years ago

a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?

Liono4ka [1.6K]

<u>Answer:</u>

The acceleration of the car is $3.13 m/s^2$

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity $15 m/s$ . It then accelerates for $4.25 s$ and in the end its velocity is $28.3 m/s$ .

initial velocity $u = 15 m/s$

time $t=4.25 s$

final velocity $v=28.3 m/s$

The equation of acceleration is

$a= \frac{(v-u)}{t}$

$= \frac{(28.3-15)}{4.25} = \frac {13.3}{4.25} =3.13m/s^2$

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

7 0

3 years ago