Answer:
X has travelled 4 times as far as Y
X is moving 2 times as fast as Y
Explanation:
Because x=1/2at²(get rid of the first portion of x=vt+1/2at² because initial v is 0 which multiplies to 0). If the acceleration is set to for example 10 (keep this constant for both X and Y) and we use t=1 and t=2 for time (as the time travelled of X is twice that of Y). Plugging this into the equation, at t=1, x=4. At t=2, x=20. 20 is four times greater than 5, so X has travelled 4 times as far as Y.
To find the difference in speed between the two objects, use the equation
. Since the initial velocity is 0, that part can just be removed from the equation. With v=at, it is easy to see that if the time plugged in is twice for one than the other (and the acceleration is the same for both), the final result will be twice of the other as well. For example: If the acceleration is 10 again for both, then v=10t. If t is 1, the velocity is 10. If t is 2, the velocity is 20.
Your answer would be A. because its talking abotu a change from being windy to raining later.
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brainly.com/question/25273589
The cooling rate of the substance is approximately 0.0732.
According to the statement, the Newton's law of cooling is defined by the following formula:
(1)
Where:
- Final temperature, in degrees Celsius.
- Initial temperature, in degrees Celsius.
- Time, in minutes.
- Cooling rate, in
.
- Current temperature, in degrees Celsius.
Please notice that substance reaches thermal equilibrium when
, that is when temperature of the substance is equal to the temperature of surrounding air.
If we know that
,
,
and
, then the cooling rate of the substance is:
![60 = 50 + (80 - 50)\cdot e^{-15\cdot k}](https://tex.z-dn.net/?f=60%20%3D%2050%20%2B%20%2880%20-%2050%29%5Ccdot%20e%5E%7B-15%5Ccdot%20k%7D)
![\frac{60-50}{80-50}= e^{-15\cdot k}](https://tex.z-dn.net/?f=%5Cfrac%7B60-50%7D%7B80-50%7D%3D%20e%5E%7B-15%5Ccdot%20k%7D)
![\frac{1}{3} = e^{-15\cdot k}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%20e%5E%7B-15%5Ccdot%20k%7D)
![k = -\frac{1}{15}\cdot \ln \frac{1}{3}](https://tex.z-dn.net/?f=k%20%3D%20-%5Cfrac%7B1%7D%7B15%7D%5Ccdot%20%5Cln%20%5Cfrac%7B1%7D%7B3%7D)
![k \approx 0.0732](https://tex.z-dn.net/?f=k%20%5Capprox%200.0732)
The cooling rate of the substance is approximately 0.0732.
To learn more on Newton's law of cooling, we kindly invite to check this verified question: brainly.com/question/13748261
Answer:
Usually at the end of your school year or the beginning of your next school year