PV=nRT
(P)(.010)=(n)(.08201)(0)
(v1/t1)=(v2/t2)
(.010/t1)=(v2/0)
The volume would be zero
B.
technically it would depend if the resistors were in series or parallel but B is the answer.
Answer:
Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.
Explanation:
Answer: F = 1235 N
Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: ![F_{net}=m.a](https://tex.z-dn.net/?f=F_%7Bnet%7D%3Dm.a)
Acceleration is the change of velocity in a period of time: ![a=\frac{\Delta v}{\Delta t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D)
Velocity of the car is in km/h. Transforming it in m/s:
![v=\frac{47.10^{3}}{36.10^{2}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B47.10%5E%7B3%7D%7D%7B36.10%5E%7B2%7D%7D)
v = 13 m/s
At the moment the car decelerates, acceleration is
a = 65 m/s²
Then, force will be
![F_{net}=19(65)](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D19%2865%29)
= 1235 N
The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.
Explanation:
Let
are the number of turns in primary and secondary coil of the transformer such that,
![\dfrac{N_p}{N_s}=\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_p%7D%7BN_s%7D%3D%5Cdfrac%7B1%7D%7B3%7D)
A resistor R connected to the secondary dissipates a power ![P_s=100\ W](https://tex.z-dn.net/?f=P_s%3D100%5C%20W)
For a transformer, ![\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_s%7D%7BN_p%7D%3D%5Cdfrac%7BV_s%7D%7BV_p%7D)
![V_s=(\dfrac{N_s}{N_p})V_p](https://tex.z-dn.net/?f=V_s%3D%28%5Cdfrac%7BN_s%7D%7BN_p%7D%29V_p)
...............(1)
The power dissipated through the secondary coil is :
![P_s=\dfrac{V_s^2}{R}](https://tex.z-dn.net/?f=P_s%3D%5Cdfrac%7BV_s%5E2%7D%7BR%7D)
![100\ W=\dfrac{V_s^2}{R}](https://tex.z-dn.net/?f=100%5C%20W%3D%5Cdfrac%7BV_s%5E2%7D%7BR%7D)
.............(2)
Let
are the new number of turns in primary and secondary coil of the transformer such that,
![\dfrac{N_p'}{N_s'}=\dfrac{1}{24}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_p%27%7D%7BN_s%27%7D%3D%5Cdfrac%7B1%7D%7B24%7D)
New voltage is :
![V_s'=(\dfrac{N_s'}{N_p'})V_p'](https://tex.z-dn.net/?f=V_s%27%3D%28%5Cdfrac%7BN_s%27%7D%7BN_p%27%7D%29V_p%27)
...............(3)
So, new power dissipated is ![P_s'](https://tex.z-dn.net/?f=P_s%27)
![P_s'=\dfrac{V_s'^2}{R}](https://tex.z-dn.net/?f=P_s%27%3D%5Cdfrac%7BV_s%27%5E2%7D%7BR%7D)
![P_s'=\dfrac{(24V_p)^2}{R}](https://tex.z-dn.net/?f=P_s%27%3D%5Cdfrac%7B%2824V_p%29%5E2%7D%7BR%7D)
![P_s'=24^2\times \dfrac{(V_p)^2}{R}](https://tex.z-dn.net/?f=P_s%27%3D24%5E2%5Ctimes%20%5Cdfrac%7B%28V_p%29%5E2%7D%7BR%7D)
![P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}](https://tex.z-dn.net/?f=P_s%27%3D24%5E2%5Ctimes%20%5Cdfrac%7B%28%5Cdfrac%7B100R%7D%7B9%7D%29%7D%7BR%7D)
![P_s'=6400\ Watts](https://tex.z-dn.net/?f=P_s%27%3D6400%5C%20Watts)
So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.