The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal chain on the scale, it read 02.1kg. The physicist, then decided to test Einstein’s equation, and compressed the spring and tied it with the chain and placed it on the scale. It read 02.7kg. Which of the following conclusions is the most likely the physicist will come to? a. Einstein's equation has an error
b. The scale is broken
c. Compressing the spring didn't add energy
d. The scale's resolution is too low to read the change in mass

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

Δm =  4.9 x 10⁻¹³ kg

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

d. The scale's resolution is too low to read the change in mass

measuring things 1 litre of water

the components of a vector are the parts of the vectors that are to each other.

b perpendicular