Answer:
699.67ft
Explanation:
We are given with,
- α = 1.2×10⁻⁵ / °C
- L₀ = 700 ft
- ΔT = -10°C − 30°C = -40°C
Now, We have to find ΔL:
- ΔL = (1.2×10⁻⁵ / °C) (700 ft) (-40°C)
- ΔL = −0.336
Rounded to two significant figures, the change in length is −0.33ft.
<u>Therefore, the final length is approximately 700 ft − 0.33 ft = 699.67ft</u>.
The height of the oil column above the water in the vessel is determined as 2 cm.
<h3>
Pressure of the vessel</h3>
The pressure of the vessel due to water, oil and silver poured into the vessel is determined from mercury column.
let level of mercury = 20 cm + 0.5 cm = 20.5 cm
20.5 cmHg = 205 mmHg
1 mmHg = 133.32 Pa
205 mmHg = 27,330.6 Pa
<h3>Height of the liquids in the vessel</h3>
P = ρgh
where;
ρ is the density of water, oil and silver respectively
ρ = 1000 kg/m³ + 881 kg/m³ + 10,800 kg/m³ = 12,681 kg/m³
h = P/(ρg)
h = (27,330.6) / (12,681 x 9.8)
h = 0.22 m
h = 22 cm
<h3>Height of oil column</h3>
Oil is less dense than water and will float on water.
Height of oil column = 22 cm - 20 cm = 2 cm
Learn more about density here: brainly.com/question/6838128
#SPJ1
Answer:
D. 5.7 m
Explanation:
The acceleration of the car is -8.7 m/s² and the car has an initial speed (u) of 10 m/s. The car said to come to rest, that means that the final velocity (v) of the car is 0 m/s.
To find the distance traveled by the car (s) before complete stop, this equation is being used:
v² = u² + 2as
The car traveled 5.7 m before coming to a complete stop
To solve this problem we will apply the concepts related to the principle of destructive and constructive interference. Mathematically this expression can be given as
Here,
n = Index of refraction
t = Thickness
m = Order of the reflection
= Wavelength
We have all of this values, therefore replacing,
Therefore the thickness of the oil slick is 233nm
The work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.
<h3>What is work done?</h3>
Work done is equal to product of force applied and distance moved.
Work = Force x Distance
Given is a block with a weight of 620 N is pulled up at a constant speed on a very smooth ramp by a constant force. The angle of the ramp with respect to the horizontal is θ = 23.5° and the length of the ramp is l = 14.1 m.
From the Newton's law of motion,
ma =F-mg sinθ =0
So, the force F = mg sinθ
Plug the values, we get
F = 620N x sin 23.5°
F = 247.224 N
Work done by motor is W= F x d
The force is equal to the weight F = mg
So, W = 247.224 x 14.1
W = 3.486 kJ
Thus, the work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.
Learn more about Work done
brainly.com/question/13662169
#SPJ1
