An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of 15°C. Both rods have the same diameter. The rods are now joined end-to-end to form a rod 90.0 cm long. If the temperature is now raised from 15°C to 90°C, what is the increase in the length of the joined rod? The coefficient of linear expansion of aluminum is 2.4 × 10-5 K-1 and that of steel is 1.2 × 10-5 K-1.
Answers
Answer:
0.9 cm
Explanation:
The computation in the increase in the length of the joined rod is shown below:
As we know that
Increase in length = increase in the length of aluminum rod + increase in The length of steel rod
= 0.9 cm
We simply added the length of aluminium rod and length of steel rod so that the length of the joined rod could come and the same is to be considered
Final answer:
The increase in length of the joined rod when the temperature is raised from 15°C to 90°C is 0.090 cm.
Explanation:
To determine the increase in length of the joined rod when the temperature is raised from 15°C to 90°C, we need to use the formula for linear expansion: AL = aLAT, where AL is the change in length, AT is the change in temperature, and a is the coefficient of linear expansion. First, we need to calculate the change in temperature for each rod: ΔT = 90°C - 15°C = 75°C. For the aluminum rod, using a linear expansion coefficient of 2.4 × 10-5 K-1 and a length of 10.0 cm, we can calculate the change in length using the formula: ALaluminum = (2.4 × 10-5 K-1)(10.0 cm)(75°C) = 0.018 cm. Similarly, for the steel rod, using a linear expansion coefficient of 1.2 × 10-5 K-1 and a length of 80.0 cm, we can calculate the change in length: ALsteel = (1.2 × 10-5 K-1)(80.0 cm)(75°C) = 0.072 cm. Since the rods are joined end-to-end, the total change in length of the joined rod is the sum of the individual changes: ΔL = ALaluminum + ALsteel = 0.018 cm + 0.072 cm = 0.090 cm. Therefore, the increase in the length of the joined rod is 0.090 cm.
Learn more about linear expansion here:
#SPJ3
Related Questions
How do you perceive the importance of online safety security ethics and etiquette
You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine the acceleration of the dragster. Would you use a. a) Linear equationb) Quadratic equationc) cubic equationd) a higher order equation
An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places
A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?
Answers
Answers
Answer:
I am sure it is A because no chemical change occurs and it is a physical change. If you can Brainllest than that would be great but if you wanna you don't have to. Hope this helps!! If wrong sorry.
Explanation:
Answers
Answer:
The maximum spring compression = 3.21 m
Explanation:
The height of the circus performer above the platform connected to string material = 5.8 m
Let the maximum compression of the spring from the impact of the circus performer be x.
According to the law of conservation of energy, the difference in potential energy of the circus performer between the initial height and the level at which spring is compressed to is equal to the work done on the spring to compress it by x
Workdone on the spring by the circus performer = (1/2)kx²
where k = spring constant = 1200 N/m
Workdone on the spring by the circus performer = (1/2)(1200)x² = 600x²
The change in potential energy of the circus performer = mg (5.8 + x)
m = mass of the circus performer = 70 kg
g = acceleration due to gravity = 9.8 m/s²
The change in potential energy of the circus performer = (70)(9.8)(5.8 + x) = (3978.8 + 686x)
600x² = 3978.8 + 686x
600x² - 686x - 3978.8 = 0
Solving this quadratic equation
x = 3.21 m or - 2.07 m
Since the negative answer doesn't satisfy the laws of physics, our correct answer is 3.21 m
Hope this Helps!!!
1. Electric field lines are the same thing as electric field vectors.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space.
3. The number of electric field lines that start or end at a charged particle is proportional to the amount of charge on the particle.
4. The electric field is strongest where the electric field lines are close together.
Answers
Answer:
All statement are correct.
Explanation:
1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.
3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.
4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.
Hence we can say that all the statement are correct.
Answers
Answer:
They would draw a total of 21.25 amperes
Explanation:
The total power consumed is
1200 W+ 750 W + 600 W= 2550 Watts
The formula relating the power consumed, the voltage and the current is given as
---------------1
given that the voltage supply is 120V
Answers
Answer:
The gel that is applied before ultrasonic imaging is a conducting material. It acts as a medium between transducer and skin. The ultrasonic waves easily transmit from the probe to the tissues because of gel. A tight bond is created between the probe and skin layer and the gel acts as a coupling agent. The density of the gel is similar to the skin layer. This reduces the attenuation of the waves. A thin layer of gel is applied which fills the air gaps and helps in transmission of waves to the tissues. Hence, the technician apply ultrasound gel to the patient before beginning the examination
The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection.