All categories

3 years ago

Which sections of the heating curve illustrate this process?

Physics

1 answer:

kirill115 [55]3 years ago

5 0

Answer:

B followed by D

Explanation:

The heat energy absorbed at B goes into potential energy that breaks the inter-molecular bonds and thus the constant temperature. Once the molecules have gained enough energy they escape the closely bonded structure and thus are free to move in random directions due to high kinetic energy. At this point (part D) an increase in heat energy leads to an increase in the kinetic energy leading to an increase in the temperature.

You might be interested in

A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:

AfilCa [17]

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

Learn more here: brainly.com/question/19072061

8 0

3 years ago

17. (16 pts) You set up a two-slit experiment with a laser that produces light with a

Natali [406]

when the two waves interfere with eachother to make a dark spot the periodic difference of the two waves is π . the wave length for 2π is 600nm

. ie. for π difference it is 300nm

4 0

4 years ago

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0

3 years ago

The wave length of violet light rounded to the nearest nanometer is a __ nm

nataly862011 [7]

Wavelength= speed / frequency

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7)

in nanometer = 413 nm

6 0

3 years ago

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t

inna [77]

Answer:

Explanation:

a. The equation of Lorentz transformations is given by:

x = γ(x' + ut')

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0

t' = 5.00 s

u =0.800 c,

c is the speed of light = 3×10⁸ m/s

Then,

γ = 1 / √ (1 - (u/c)²)

γ = 1 / √ (1 - (0.8c/c)²)

γ = 1 / √ (1 - (0.8)²)

γ = 1 / √ (1 - 0.64)

γ = 1 / √0.36

γ = 1 / 0.6

γ = 1.67

Therefore, x = γ(x' + ut')

x = 1.67(0 + 0.8c×5)

x = 1.67 × (0+4c)

x = 1.67 × 4c

x = 1.67 × 4 × 3×10⁸

x = 2.004 × 10^9 m

x ≈ 2 × 10^9 m

Now, to find t we apply the same analysis:

but as x'=0 we just have:

t = γ(t' + ux'/c²)

t = γ•t'

t = 1.67 × 5

t = 8.35 seconds

b. Mavis reads 5 s on her watch which is the proper time.

Stanley measured the events at a time interval longer than ∆to by γ,

such that

∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)

c. According to Stanley,

dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m

which is the same as in part (a)

7 0

3 years ago