So, the total energy value that has been transferred by work is <u>32 kJ</u>.
<h3>Introduction</h3>
Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :
![\boxed{\sf{\bold{W = F \times s}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7BW%20%3D%20F%20%5Ctimes%20s%7D%7D%7D%20)
With the following condition :
- W = work (J)
- F = force (N)
- s = shift or displacement (m)
Now, because in this question, the "s" is not directly known, whereas it is known that the initial velocity is zero, the object has an acceleration, and is moving in certain time intervals. Then, use this formula to find the value of "s" !
![\boxed{\sf{\bold{s = \frac{1}{2} \times a \times t^2}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7Bs%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20a%20%5Ctimes%20t%5E2%7D%7D%7D%20)
With the following condition :
- s = shift or displacement (m)
- a = acceleration (m/s²)
- t = interval of the time (s)
<h3>Problem Solving </h3>
We know that :
- a = acceleration = 4 m/s²
- t = interval of the time = 20 s
- m = mass = 10 kg
What was asked :
Step by Step :
- Calculate value of force by multiple m and a
![\sf{F = m \times a}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%20m%20%5Ctimes%20a%7D%20)
![\sf{F = 10 \times 4}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%2010%20%5Ctimes%204%7D%20)
![\sf{\bold{F = 40 \: N}}](https://tex.z-dn.net/?f=%20%5Csf%7B%5Cbold%7BF%20%3D%2040%20%5C%3A%20N%7D%7D%20)
- Calculate value of shift or displacement
![\sf{s = \frac{1}{2} \times a \times t^2}](https://tex.z-dn.net/?f=%20%5Csf%7Bs%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20a%20%5Ctimes%20t%5E2%7D%20)
![\sf{s = \frac{1}{\cancel 2} \times \cancel 4 \:_2 \times 20^2}](https://tex.z-dn.net/?f=%20%5Csf%7Bs%20%3D%20%5Cfrac%7B1%7D%7B%5Ccancel%202%7D%20%5Ctimes%20%5Ccancel%204%20%5C%3A_2%20%5Ctimes%2020%5E2%7D%20)
![\sf{s = 2 \times 400}](https://tex.z-dn.net/?f=%20%5Csf%7Bs%20%3D%202%20%5Ctimes%20400%7D%20)
![\sf{\bold{s = 800 \: m}}](https://tex.z-dn.net/?f=%20%5Csf%7B%5Cbold%7Bs%20%3D%20800%20%5C%3A%20m%7D%7D%20)
- Last, calculate the value of work
![\sf{W = F \times s}](https://tex.z-dn.net/?f=%20%5Csf%7BW%20%3D%20F%20%5Ctimes%20s%7D%20)
![\sf{W = 40 \times 800}](https://tex.z-dn.net/?f=%20%5Csf%7BW%20%3D%2040%20%5Ctimes%20800%7D%20)
![\boxed{\sf{W = 32,000 \: J = 32 \: kJ}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7BW%20%3D%2032%2C000%20%5C%3A%20J%20%3D%2032%20%5C%3A%20kJ%7D%7D%20)
So, the total energy value that has been transferred by work is 32 kJ.
Answer:
Time taken = 10400 s
Explanation:
Given:
Initial speed of the train, ![u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}](https://tex.z-dn.net/?f=u%3D72%5Ctextrm%7B%20km%2Fh%7D%3D72%5Ctimes%20%5Cfrac%7B5%7D%7B18%7D%3D20%5Ctextrm%7B%20m%2Fs%7D)
Final speed of the train, ![v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}](https://tex.z-dn.net/?f=v%3D90%5Ctextrm%7B%20km%2Fh%7D%3D90%5Ctimes%20%5Cfrac%7B5%7D%7B18%7D%3D25%5Ctextrm%7B%20m%2Fs%7D)
Displacement of the train, ![S=234\textrm{ km}=234\times 1000=234000\textrm{ m}](https://tex.z-dn.net/?f=S%3D234%5Ctextrm%7B%20km%7D%3D234%5Ctimes%201000%3D234000%5Ctextrm%7B%20m%7D)
Using Newton's equation of motion,
![v - u = at\\a=\frac{v-u}{t}](https://tex.z-dn.net/?f=%20v%20-%20u%20%3D%20at%5C%5Ca%3D%5Cfrac%7Bv-u%7D%7Bt%7D)
Now, using Newton's equation of motion for displacement,
![v^{2}-u^{2}=2aS](https://tex.z-dn.net/?f=v%5E%7B2%7D-u%5E%7B2%7D%3D2aS)
Now, plug in the value of
in the above equation. This gives,
![v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}](https://tex.z-dn.net/?f=v%5E%7B2%7D-u%5E%7B2%7D%3D2%5Ctimes%20%5Cfrac%7Bv-u%7D%7Bt%7D%5Ctimes%20S%5C%5C%28v%2Bu%29%28v-u%29%3D%5Cfrac%7B2%28v-u%29S%7D%7Bt%7D%5C%5Ct%3D%5Cfrac%7B2%28v-u%29S%7D%7B%28v%2Bu%29%28v-u%29%7D%5C%5Ct%3D%5Cfrac%7B2S%7D%7Bv%2Bu%7D)
Now, plug in 234000 m for
, 25 m/s for
and 20 m/s for
. Solve for
.
![t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2S%7D%7Bv%2Bu%7D%5C%5Ct%3D%5Cfrac%7B2%5Ctimes%20234000%7D%7B25%2B20%7D%5C%5Ct%3D%5Cfrac%7B468000%7D%7B45%7D%3D10400%5Ctextrm%7B%20s%7D)
Therefore, the time taken by the train is 10400 s.
Answer: Erosion or water level rising
Explanation: Erosion happens when water goes against a rock and breaks it down little by little so these mountains may have once been above water but became eroded. Also, people might’ve added so much pollution into the ocean that the water level rose or there was an increase in the population of large animals. Sorry for my very long answer lol
Answer:
Time, t = 0.104 seconds
Explanation:
Frequency of the click of the Dolphin, f = 55.3 kHz
A dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below, d = 80 m
The speed of sound in seawater is, v = 1530 m/s
Once the sound is send and reflects, the total distance covered by it is 2d such that,
![t=\dfrac{2d}{v}\\\\t=\dfrac{2\times 80}{1530}\\\\t=0.104\ s](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B2d%7D%7Bv%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B2%5Ctimes%2080%7D%7B1530%7D%5C%5C%5C%5Ct%3D0.104%5C%20s)
So, the time elapses before the dolphin hears the echoes of the clicks is 0.104 seconds.
a)5m/s b)5
the 5 is because you add the seconds to get 8 seconds and then do the same with the distance to get 40. 40/8 = 5. speed = 5
Velocity = displacement/change in time
V = 40/8
I just realized how unorganised my math looks but I hope this is helpfull