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3 years ago

in the diagram the solid eventually becomes a gas what will happen to bring the substance from a solid to a gas

1 answer:

8 0

Since there is no diagram I can see to prove my answer, I must infer that a heat source should be applied to the solid substance and cause the atoms in the solid to spread further apart and become a different state, in which atoms will be able to freely move, gas.

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1) how does reflection differ from diffraction?

Evgesh-ka [11]

Answer: (1) The correct answer is A.

(2) The correct answer is D.

Explanation:

(1)

Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.

Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.

Therefore, the correct answer is A.

(2)

Amplitude is the maximum displacement in the medium from the rest position.

The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.

Therefore, the correct option is D.

5 0

3 years ago

Read 2 more answers

A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect fricti

topjm [15]

Answer:

1.66 m/s

Explanation:

Work or kinetic energy = $\frac{1}{2} mv^{2}$

$4864=\frac{1}{2} (3540)v^{2}$

v = 1.66 m/s

6 0

3 years ago

Questions<br> 2.<br> Rewrite the following quantities using suitable<br> prefixes<br><br> 5000 000

Rufina [12.5K]

Ans. 5 X 10^6

Explanation:

Here,^ represents 6 times 10 and 5 is multiplied. That is 5000 000.

Hope this helps

5 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

a rocket with a mass of 4kg accelerates up from the ground at a rate of 20ft m/s^2, what is the drag force acting on the rocket

Maru [420]

The drag force acting on the rocket is 80N.

<h3>Give an explanation of drag force?</h3>

The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.

The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.

The drag force acting on the rocket F = ma

Given,

m = 4kg, a = 20ftm/s²

Substituting m and a values in the above formula,

The drag force acting on the rocket F = 4×20

The drag force acting on the rocket F = 80N.

To know more about drag force visit:

brainly.com/question/15144984

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8 0

1 year ago