A brass measuring rule is connected 15°c the volume obtain is measured with the rule at 35°c appear to be 841.4cm. What is the length of the object? (Linear expansivity of brass=1.8×10-5 k-1)

new length= original length*(1+1.8e-5*20)

original length= 841.4/(1.00036)
=841.097205 change to significant digits given

I need the answer

A brass cube has a volume of 100cm3 at 25 degrees Celsius calculate it's volume at 0 degree Celsius linear expansivity of brass equals 2.0×10-5k-1

Well, well, well, we have ourselves a hot measuring rule! So, let's solve this with a touch of humor, shall we?

First things first, we need to find the coefficient of linear expansion for brass. You mentioned it's 1.8×10-5 k-1, but I assure you, it hasn't received any formal clown training.

Now, using the formula for linear expansion, which is ΔL = α×L×ΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.

In this case, we want to find the original length of the object, so we'll rearrange the equation to solve for L: L = ΔL / (α×ΔT).

Plugging in the values, we have ΔL = 841.4 cm (that's quite the expansion!), α = 1.8×10-5 k-1, and ΔT = 35°C - 15°C = 20°C.

Now, let the hilarious calculations begin:

L = 841.4 cm / (1.8×10-5 k-1 × 20°C)
L = 841.4 / (3.6×10-4 k)
L ≈ 2337222.22 cm

Voilà! The length of the object is approximately 2,337,222.22 cm. Now that's what I call a long measure! Hope that brings a smile to your face!

To solve this problem, we can use the concept of thermal expansion. The formula for linear expansion is given by:

ΔL = L₀ * α * ΔT

Where:
ΔL is the change in length
L₀ is the original length
α is the coefficient of linear expansivity
ΔT is the change in temperature

In this case, we are given that the linear expansivity of brass is 1.8×10^(-5) K^(-1). We are also given that the temperature changes from 15°C to 35°C.

Let's assume the original length of the object is L.

First, we need to calculate the change in temperature:
ΔT = T₂ - T₁
ΔT = 35°C - 15°C
ΔT = 20°C

Now, we can calculate the change in length using the formula:

ΔL = L₀ * α * ΔT
ΔL = L * (1.8×10^(-5) K^(-1)) * 20°C

Next, we are given that the volume obtained is measured with the rule at 35°C and is equal to 841.4 cm^3. Since the object is assumed to be a rectangular shape, we can use the formula for volume:

V = L * W * H

Since the width (W) and height (H) are constant, we can assume the change in volume is equivalent to the change in length (ΔL).

Therefore, the volume change is:

ΔV = ΔL = L * (1.8×10^(-5) K^(-1)) * 20°C

Finally, we can calculate the length of the object (L) using the given volume change:

L = ΔV / (α * ΔT)
L = (L * (1.8×10^(-5) K^(-1)) * 20°C) / (1.8×10^(-5) K^(-1) * 20°C)

Simplifying the equation:

L = L / L

Therefore, the length of the object is equal to 1.

So, the length of the object is 1 unit (the actual units used for L were not provided).