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Crazy boy [7]
3 years ago
8

2.Which choice most accurately explains what happens to the bonds between atoms when water changes to steam, and what becomes of

the energy added during this phase change?
When water changes to steam bonds between molecules are created. After reaching the boiling point, energy is used to change bond strength and change the kinetic energy of the atoms.
When water changes to steam bonds between molecules break apart. After reaching the boiling point, energy is used to break bonds, but does not affect the kinetic energy of the atoms.
When water changes to steam bonds between molecules are created. After reaching the boiling point, energy is used to break bonds, but does not affect the kinetic energy of the atoms.
When water changes to steam bonds between molecules break apart. After reaching the boiling point, energy is used to change bond strength and change the kinetic energy of the atoms.
Physics
2 answers:
erastova [34]3 years ago
6 0

Answer:

Explanation:

The conversion of water into steam involves various factors and changes.

When heat is applied to water, the bond between the molecules start to loosen up and molecules starts separating hence it from changes from liquid to gas or steam.

At this time molecules have greater kinetic energy which is given to them in form of heat. Hence apart from all options given, D is the correct one which states that :

"When water changes to steam bonds between molecules break apart. After reaching the boiling point, energy is used to change bond strength and change the kinetic energy of the atoms."

seraphim [82]3 years ago
5 0

Answer:

When water turns to steam,

D) the water absorbs energy which causes the water molecules to have more kinetic and potential energy changing their configuration from a liquid to a gas.

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A 0.0780 kg lemming runs off a
kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0
3 years ago
a nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm. how many photons are in the pulse?
Wewaii [24]

A nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm and has 1785 x 10¹⁹ photons in the pulse.

<h3>How many photons are in the pulse?</h3>

Energy of a single photon is

E=hcλ

E=6.626×10⁻³⁴ J s×3×108 m/s /340×10⁻⁹ m

E=6.31×10⁻¹⁹  J

Number of photons in the laser is

n=Total Energy/Energy per photon

n=10⁷×10⁻³J /5.90×10⁻¹⁹J/photon

n= 1785 x 10¹⁹ photons

To learn about photons, refer: brainly.com/question/20912241?referrer=searchResults

#SPJ4

6 0
2 years ago
What is the name given to a material with zero resistance that can conduct electricity without a loss of energy?
tatyana61 [14]
It would be C. Superconductor 
3 0
3 years ago
A car turns a certain curve of radius 24.98 m with constant linear speed of
Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

 = 3525.19 kg

7 0
3 years ago
You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

\Delta E_k=W=W_g-W_p

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J

hence, the change in Er is about 1.52J times the initial rotational energy

5 0
3 years ago
Read 2 more answers
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