1) 15 / 12 = 1.25 ratio
2) to increase acceleration 1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg
choose A
Answer:
Explanation:
In case of gas , work done
W = ∫ p dV , p is pressure and dV is small change in volume
the limit of integration is from Vi to Vf .
= ∫ p dV
= ∫ p₀
dV
= p₀
/ (
)
= - 5p₀ 
Taking limit from Vi to Vf
W = - 5 p₀ (
) ltr- atm.
Answer:
The speed in the first point is: 4.98m/s
The acceleration is: 1.67m/s^2
The prior distance from the first point is: 7.42m
Explanation:
For part a and b:
We have a system with two equations and two variables.
We have these data:
X = distance = 60m
t = time = 6.0s
Sf = Final speed = 15m/s
And We need to find:
So = Inicial speed
a = aceleration
We are going to use these equation:


We are going to put our data:


With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.



![[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}](https://tex.z-dn.net/?f=%5B%5Csqrt%7B%2815m%2Fs%29%5E2-%282%2Aa%2A60m%29%7D%5D%5E%7B2%7D%3D%5B15m%2Fs-%28a%2A6s%29%5D%5E%7B2%7D)








If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2
After, we are going to determine the speed in the first point:




For part c:
We are going to use:




Answer
The rate at which the magnetic field is changing is
Explanation
From the question we are told that
The electric field strength is 
The radius is 
The rate of change of the magnetic field is mathematically represented as

Where
is change of a unit length

Where A is the area which is mathematically represented as

So
where L is the circumference of the circle which is mathematically represented as

So
![E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]](https://tex.z-dn.net/?f=E%20%282%20%5Cpi%20r%20%29%20%3D%20%20%28%5Cpi%20r%5E2%20%29%20%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D)
![E = \frac{r}{2} [\frac{dB}{dt} ]](https://tex.z-dn.net/?f=E%20%20%3D%20%20%20%5Cfrac%7Br%7D%7B2%7D%20%20%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D)
![[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D%20%3D%20%5Cfrac%7BE%7D%7B%20%5Cfrac%7Br%7D%7B2%7D%20%7D)
substituting values
![[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D%20%3D%20%5Cfrac%7B3.5%20%2A10%5E%7B-3%7D%7D%7B%20%5Cfrac%7B15%7D%7B2%7D%20%7D)
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