<span>The
kinetic energy is the work done by the object due to its motion. It is
represented by the formula of the half the velocity squared multiply by the
mass of the object. In this problem, you have two vehicles, the other one is large and the
other one is small. Let us assume that they travel with the same velocity. Note
that the kinetic energy is proportional to the mass of the object. So when you
increase the mass of the other, it also increases the kinetic energy of that
object. The same holds true for the two vehicles. The larger the vehicle, its
kinetic energy is also large and therefore its stopping distance will be longer
than that of the smaller vehicle.</span>
Answer:
Explanation:
24 - gauge wire , diameter = .51 mm .
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm
= .084 ohm
B ) Current required through this wire
= 12 / .084 A
= 142.85 A
C )
Let required length be l
resistance = .084 l
2 = 12 / .084 l
l = 12 / (2 x .084)
= 71.42 m
Answer:
option (d)
Explanation:
The relation between the rms velocity and the molecular mass is given by
v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant
So for two gases
![\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7BA%7D%7D%7Bv_%7BB%7D%7D%3D%5Csqrt%7B%5Cfrac%7BM_%7BB%7D%7D%7BM_%7BA%7D%7D%7D)
![\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}](https://tex.z-dn.net/?f=%5Cfrac%7B2v_%7BB%7D%7D%7Bv_%7BB%7D%7D%3D%5Csqrt%7B%5Cfrac%7BM_%7BB%7D%7D%7BM_%7BA%7D%7D%7D)
![{\frac{M_{B}}{M_{A}}} = 4](https://tex.z-dn.net/?f=%7B%5Cfrac%7BM_%7BB%7D%7D%7BM_%7BA%7D%7D%7D%20%3D%204)
![{\frac{M_{B}}{4}} = M_{A}](https://tex.z-dn.net/?f=%7B%5Cfrac%7BM_%7BB%7D%7D%7B4%7D%7D%20%3D%20M_%7BA%7D)
Answer:
The taken is ![t_A = 19.0 \ s](https://tex.z-dn.net/?f=t_A%20%20%3D%2019.0%20%5C%20s)
Explanation:
Frm the question we are told that
The speed of car A is ![v_A = 22 \ m/s](https://tex.z-dn.net/?f=v_A%20%20%3D%20%2022%20%5C%20m%2Fs)
The speed of car B is ![v_B = 29.0 \ m/s](https://tex.z-dn.net/?f=v_B%20%20%3D%2029.0%20%5C%20m%2Fs)
The distance of car B from A is ![d = 300 \ m](https://tex.z-dn.net/?f=d%20%3D%20300%20%5C%20m)
The acceleration of car A is ![a_A = 2.40 \ m/s^2](https://tex.z-dn.net/?f=a_A%20%20%3D%202.40%20%5C%20m%2Fs%5E2)
For A to overtake B
The distance traveled by car B = The distance traveled by car A - 300m
Now the this distance traveled by car B before it is overtaken by A is
![d = v_B * t_A](https://tex.z-dn.net/?f=d%20%3D%20v_B%20%2A%20t_A)
Where
is the time taken by car B
Now this can also be represented as using equation of motion as
![d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300](https://tex.z-dn.net/?f=d%20%3D%20v_A%20t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_A%20t_A%5E2%20-%20300)
Now substituting values
![d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=d%20%3D%2022t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
Equating the both d
![v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=v_B%20%2A%20t_A%20%3D%2022t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
substituting values
![29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=29%20%2A%20t_A%20%3D%2022t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
![7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=7%20t_A%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
![7 t_A =1.2 t_A^2 - 300](https://tex.z-dn.net/?f=7%20t_A%20%3D1.2%20t_A%5E2%20-%20300)
![1.2 t_A^2 - 7 t_A - 300 = 0](https://tex.z-dn.net/?f=1.2%20t_A%5E2%20-%207%20t_A%20-%20300%20%20%3D%200)
Solving this using quadratic formula we have that
![t_A = 19.0 \ s](https://tex.z-dn.net/?f=t_A%20%20%3D%2019.0%20%5C%20s)