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3 years ago

Select the correct answer.

Physics

2 answers:

Lemur [1.5K]3 years ago

8 0

Answer:

B.Temperature and density increase with layer thickness.

Explanation:

Serhud [2]3 years ago

5 0

Answer:

b hope this helps

Explanation:

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A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What vol

Gwar [14]

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

$C=\frac{n}{V}$

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

$230mL=0.23L$

$n=C*V=0.240 M*0.23L=0.055 mol$

Now, we isolate the variable V to know the new volume with the new concentration given.

$V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL$

Finally, the volume of water evaporated is the difference between initial and final volume.

$V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL$

4 0

3 years ago

A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a

mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required $t=\frac{distance}{speed}=\frac{d}{38}hour$

Speed when car return from hill = 66 km/h

So time required to return fro hill $t=\frac{d}{66}h$

Total time $t_{total}=\frac{t}{38}+\frac{t}{66}$

Total distance = d+d =2d

So average speed $=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h$

8 0

3 years ago

Natalie accelerate her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 s. find her average acceleration

Vikentia [17]

Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s

acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²

Answer: the acceleration would be 1.6 m/s²

8 0

3 years ago

A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos

faust18 [17]

Answer:

The possible thickness of the soap bubble = $1.034\times 10^{-7}\ m.$

Explanation:

<u>Given:</u>

Refractive index of the soap bubble, $\mu=1.33.$
Wavelength of the light taken, $\lambda = 550.0\ nm = 550.0\times 10^{-9}\ m.$

Let the thickness of the soap bubble be $t$ .

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

$2\mu t=\left ( m+\dfrac 12 \right )\lambda.$

where $m$ is the order of constructive interference.

Since the soap bubble is appearing very bright, the order should be 0, as $0^{th}$ order interference has maximum intensity.

Thus,

$2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.$

It is the possible thickness of the soap bubble.

6 0

3 years ago

(a) a light-rail commuter train accelerates at a rate of 1.15 m/s2. how long does it take it to reach its top speed of 80.0 km/h

snow_lady [41]

A = 1.15m/s2, Vf = 80.0km/h --> we need it in m/s, so:
Vf = 80km/h × 1000m/1km × 1h/3600s
= 22.22m/s
Top speed = Vf, initial speed = Vi
time (t) = V(Vf-Vi) ÷ a
t = (22.22-0)m/s ÷ 1.15m/s2
t = 22.22m/s × s2/1.15m
= 19.32 seconds

6 0

3 years ago