First, solve for the acceleration of the car. You know the mass of the car and the braking force, so you can use the equation Force = Mass x Acceleration. This gives you 12,000 = 2,000 x A. Divide 12,000 by 2,000 to find the acceleration equal to 6 m/s^2. This is the rate that the car is slowing down at. Velocity is equal to accleration x time (rate x time), so you multiply 6 by the time of 5 seconds. This leaves you with a velocity of 30 m/s or about 67.1 mph.
Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law
![V = iR](https://tex.z-dn.net/?f=V%20%3D%20iR)
![V = (2 mA)(10 kohm)](https://tex.z-dn.net/?f=V%20%3D%20%282%20mA%29%2810%20kohm%29)
![V = 20 volts](https://tex.z-dn.net/?f=V%20%3D%2020%20volts)
so voltage across the capacitor + voltage across resistor = V
![V_c + 20 = 50](https://tex.z-dn.net/?f=V_c%20%2B%2020%20%3D%2050)
![V_c = 30 V](https://tex.z-dn.net/?f=V_c%20%3D%2030%20V)
Now we know that
![U = \frac{q^2}{2C}](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7Bq%5E2%7D%7B2C%7D)
here rate of change in energy of the capacitor is given as
![\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdU%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bq%7D%7BC%7D%20%5Cfrac%7Bdq%7D%7Bdt%7D)
![\frac{dU}{dt} = (30)(2 mA)](https://tex.z-dn.net/?f=%5Cfrac%7BdU%7D%7Bdt%7D%20%3D%20%2830%29%282%20mA%29)
![\frac{dU}{dt} = 0.06 W](https://tex.z-dn.net/?f=%5Cfrac%7BdU%7D%7Bdt%7D%20%3D%200.06%20W)
The formula for velocity vf = vi + at
First list your given information
2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)
Since you want the a for acceleration get a by itself
a = (vf-vi)/t
So a= (6-2)/2
a= 4/2
a=2
Now units
the units for acceleration are m/s
![x^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20)
2m/s
Answer:
Part a)
![\tau = 23.1 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%2023.1%20Nm)
Part b)
![\tau = 17.05 Foot pound force](https://tex.z-dn.net/?f=%5Ctau%20%3D%2017.05%20Foot%20pound%20force)
Explanation:
As we know that torque is defined as the product of force and its perpendicular distance from reference point
so here we have
![\tau = \vec r \times \vec F](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cvec%20r%20%5Ctimes%20%5Cvec%20F)
now we have
![\tau = (0.140)(165)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%280.140%29%28165%29)
![\tau = 23.1 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%2023.1%20Nm)
Part b)
Now we know the conversion as
1 meter = 3.28 foot
1 N = 0.225 Lb force
now we have
![\tau = 23.1 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%2023.1%20Nm)
![\tau = 23.1 (0.225 Lb)(3.28 foot)](https://tex.z-dn.net/?f=%5Ctau%20%3D%2023.1%20%280.225%20Lb%29%283.28%20foot%29)
![\tau = 17.05 Foot pound force](https://tex.z-dn.net/?f=%5Ctau%20%3D%2017.05%20Foot%20pound%20force)