Answer:
the density makes something float or sink below water its the "water weight "
Explanation:
According to Ptolemy's model, he, too, believed in a geocentric Universe and that the planets and stars were perfect spheres, though Earth itself was not.
He further thought that the movements of the planets and stars must be circular since they were perfect and, if the motions were circular, then they could go on forever.
<h3>What is comets and shooting stars?</h3>
Shooting stars are very different from comets, although the two can be related. A Comet is a ball of ice and dirt, orbiting the Sun (usually millions of miles from Earth). ... A shooting star on the other hand, is a grain of dust or rock (see where this is going) that burns up as it enters the Earth's atmosphere.
Learn more about ptolemy's model:
brainly.com/question/12639459
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
Answer:
The answer to your question is 5.4 cm
Explanation:
This problem refers to calculate the change in length in one dimension due to a change in temperature.
Data
α = 12 x 10⁻⁶
Lo = 150 meters
ΔT = 30 °C
Formula
ΔL/Lo = αΔT
solve for ΔL
ΔL = αLoΔT
Substitution
ΔL = (12 x 10⁻⁶)(150)(30)
Simplification
ΔL = 0054 m = 5.4 cm
