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OLga [1]
2 years ago
12

What happens to the chemical structure of water when it changes state?

Physics
1 answer:
Marta_Voda [28]2 years ago
5 0

Answer:

c) Water molecules melt into gas molecules.

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A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

E_{A}=E_{B}

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

E_{B}=m*g*h+\frac{1}{2}*m*v^{2}

Therefore we will have the following equation:

(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]

5 0
3 years ago
a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical
nlexa [21]

Answer:

Vx =  10.9 m/s ,  Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s  × cos 55°

Vx =  10.9 m/s

Vx = V sin θ

Vy = 19 m/s  × sin 55°

Vy = 15.6 m/s

6 0
3 years ago
Read 2 more answers
What is the speed of an 800 kg automobile if it has a kinetic energy of 9.00 x 10^J?
MA_775_DIABLO [31]

Ek = 1/2 mv^2

9 × 10^4 = 1/2 × 800 × v^2

9 × 10^4/400 = 400 v^2 / 400

9 × 10^4/400 = v^2

√225 = v

15 ms⁻¹ = v

That's the only way I know how to work it out

I think in this case velocity and speed would be considered the same because me

s = d/t and v=d/t

one is distance travelled and the other is displacement of a body

7 0
2 years ago
-poopp poopoop jujuj
san4es73 [151]

Answer:

okay.........

Explanation:

3 0
3 years ago
Read 2 more answers
If an astronaut has a mass of 80kg on earth, what is their mass on the moon?
stealth61 [152]

Answer:

130 N basically

Explanation:

6 0
2 years ago
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