Answer:
Final Length = 30 cm
Explanation:
The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:
F = kΔx
where,
F = Force applied
k = spring constant
Δx = change in length of spring
First, we find the spring constant of the spring. For this purpose, we have the following data:
F = 50 N
Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m
Therefore,
50 N = k(0.05 m)
k = 50 N/0.05 m
k = 1000 N/m
Now, we find the change in its length for F = 100 N:
100 N = (1000 N/m)Δx
Δx = (100 N)/(1000 N/m)
Δx = 0.1 m = 10 cm
but,
Δx = Final Length - Initial Length
10 cm = Final Length - 20 cm
Final Length = 10 cm + 20 cm
<u>Final Length = 30 cm</u>
Heat required to melt 0.05 kg of aluminum is 28.7 kJ.
<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>
The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.
The formula to be used is given below:
- Heat required = mass * heat capacity * temperature change
Assuming the aluminum sheet was at room temperature initially.;
Room temperature = 25 °C
Melting point of aluminum = 660.3 °C
Temperature difference = (660.3 - 25) = 635.3 903
Heat capacity of aluminum = 903 J/kg/903
Heat required = 0.05 * 903 * 635.3
Heat required = 28.7 kJ
In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.
Learn more about heat capacity at: brainly.com/question/21406849
#SPJ1
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
Answer:
Velocity
Explanation:
We finds that the winds are coming from the west at 15 miles per hour. This information shows the velocity of the wind. Since, velocity is a vector quantity. It has both magnitude and direction. 15 miles per hour shows the speed of wind and west shows the direction of wind motion.
Hence, the given information describes wind velocity.
Answer:
Crust, Upper mantle, mantle, outer core, inner core
Explanation:
The Earth's layers have been clasified in 5 according to the materials that conform them, theri physical properties, strengths and also their state of matter. We all know how the outer layer of the Earth looks like, but if we start to dig a huge hole we are going to see different types por materials due to a change in pressure, temperature, and other factors. At the very center of the Earth there's what's called "core". The core is liquid and at extremely high temperatures. This is because of the enormous amount of pressure the rest of the Earth is putting it under. So, if we list the different layers of the Earth according to the materials they are made of, from the Earth's surface to the core, the answer is:
1) Crust (surface)
2) Upper Mantle
3) Mantle
4) Outer core
5) Inner core
In some books you may find a layer called Lithosphere. Tis layer consists not only of the crust, but also it contains the transition zone between the upper mantle and the crust.
