Explanation:
The volume of the bread decreases, making the bread appear more compact, and smaller in size. The mass stays the same, it won't change unless part of the bread is removed. The density increases, the air bubbles inside of the bread get squished down, causing the bread to be smaller, and in turn, causing it to be more solid.
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Your friend in answering,
~Steve
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
I really don’t know but I think it’s D
Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is 
Explanation:
From the question we are told that
The semi - major axis of the rocky debris 
The semi - major axis of Planet D is 
The orbital period of planet D is 
Generally from Kepler third law
Here T is the orbital period while a is the semi major axis
So
=> ![T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%20T_D%20%2A%20%20%5B%5Cfrac%7Ba_R%7D%7Ba_D%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=> ![T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=>
C, they didn't know any better
