Answer:
As per Provided Information
- Mass of car m is 1400Kg
- Initial velocity u is 13m/s
- Time taken to stop t is 5 second .
- Final velocity v is 0m/s
we have been asked to determine the force applied to stop the car. First we will calculate the acceleration of the car.
![\boxed{\bf \: a \: = \dfrac{(v - u)}{t}}](https://tex.z-dn.net/?f=%20%20%5Cboxed%7B%5Cbf%20%5C%3A%20a%20%5C%3A%20%20%3D%20%20%5Cdfrac%7B%28v%20-%20u%29%7D%7Bt%7D%7D)
Substituting the value and let's solve it
![\longrightarrow\sf \: a \: = \dfrac{0 - 13}{5} \\ \\ \\ \longrightarrow\sf \: a \: = \cfrac{ - 13}{5} \\ \\ \\ \longrightarrow\sf \: a \: = - 2.6 \: {ms}^{ - 2}](https://tex.z-dn.net/?f=%20%20%5Clongrightarrow%5Csf%20%5C%3A%20a%20%5C%3A%20%20%3D%20%20%5Cdfrac%7B0%20-%2013%7D%7B5%7D%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%5Clongrightarrow%5Csf%20%5C%3A%20a%20%5C%3A%20%20%3D%20%20%5Ccfrac%7B%20-%2013%7D%7B5%7D%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%5Clongrightarrow%5Csf%20%5C%3A%20a%20%5C%3A%20%20%3D%20%20%20-%202.6%20%5C%3A%20%20%7Bms%7D%5E%7B%20-%202%7D%20)
Now, let's calculate the force applied to stop the car .
![\pink{\boxed{\bf \: F = ma}}](https://tex.z-dn.net/?f=%20%20%20%20%5Cpink%7B%5Cboxed%7B%5Cbf%20%5C%3A%20F%20%3D%20ma%7D%7D)
Substituting the value we get
![\longrightarrow \sf \: F = 1400 \times ( - 2.6) \\ \\ \\ \longrightarrow \sf \: F = - 3640 \: N](https://tex.z-dn.net/?f=%20%5Clongrightarrow%20%5Csf%20%5C%3A%20F%20%3D%201400%20%5Ctimes%20%28%20-%202.6%29%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%5Clongrightarrow%20%5Csf%20%5C%3A%20F%20%3D%20-%203640%20%5C%3A%20N)
Here , negative sign show that the " Force is acting in opposite direction of the motion"
<u>Therefore</u><u>,</u>
- <u>3</u><u>6</u><u>4</u><u>0</u><u> </u><u>Newton</u><u> </u><u>force </u><u>is </u><u> </u><u>required</u><u> </u><u>to</u><u> </u><u>stop</u><u> </u><u>the </u><u>car </u><u>.</u>
Answer:
419.25 lbs
Explanation:
Let iron density
be 491.5 lb/ft3
and aluminum density
be 169 lb/ft3
We can calculate the mass of iron that displaces 1.3ft3
![m_i = V*\rho_i = 1.3*491.5 = 638.95 lb](https://tex.z-dn.net/?f=m_i%20%3D%20V%2A%5Crho_i%20%3D%201.3%2A491.5%20%3D%20638.95%20lb)
Also the mass of aluminum that displaces 1.3 ft3
![m_a = V*\rho_a = 1.3*169 = 219.7 lb](https://tex.z-dn.net/?f=m_a%20%3D%20V%2A%5Crho_a%20%3D%201.3%2A169%20%3D%20219.7%20lb)
So by using aluminum, the airplane can save
638.95 - 219.7 = 419.25 lbs of weight
Answer: 574.59 K
Hope this helps you out (◠‿◕)
Answer:
Because of the formula ![E=mc^2](https://tex.z-dn.net/?f=E%3Dmc%5E2)
Explanation:
In this problem we are describing two different processes:
- Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei
- Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus
In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.
According to Einsten's formula, this mass difference has been converted into energy, as follows:
![E=\Delta mc^2](https://tex.z-dn.net/?f=E%3D%5CDelta%20mc%5E2)
where:
E is the energy released in the reaction
is the mass defect, the difference between the final total mass and the initial total mass
is the speed of light
From the formula, we see that the factor
is a very large number, therefore even if the mass defect
is very small, nuclear fusion and nuclear fission release huge amounts of energy.