Physics

Vernier calipers are instruments for making accurate linear measurements. They are commonly used to measure the diameter of a sphere or the depth of a hole. Its precision comes from a sliding Vernier scale that runs along a fixed main scale. 

There are two types of errors in measurements. One is a systematic error, which comes from a fault in the measuring device. The other is a random error in measurement, which is caused by a sudden change in the environment.  

Sign conventions are important because all kinematic variables can be positive or negative. You must first choose your coordinate system and positive direction, then consistently apply signs. For example, if you are choosing upward as positive in free-fall problems, gravity becomes negative (a = -g), upward initial velocity is positive, and downward displacement is negative. The equations of motion work for any situation, as long as you substitute values with proper signs. Incorrect conventions lead to wrong answers.

It's not always true. Direction of velocity is the most important consideration here that will tell us whether acceleration increases or decreases speed. You can consider two scenarios. If you're falling, that's negative velocity. That implies negative gravitational acceleration. Here, your speed increases. If you're moving upward, that's positive velocity. Now, with that same negative acceleration, your speed decreases. For choosing the equations of motion, you need to know both the signs of acceleration and velocity to determine if you're speeding up or slowing down.

Zero velocity doesn't mean zero acceleration. When you throw a ball-like object upward, at its peak the velocity is zero. But acceleration remains constant due to gravity. The velocity is still changing from positive to negative at that instant. That means acceleration continues.

70 g – N = 70 a = 70 * 0.2 = 14

->N = 700 – 14 = 686 N

$F=\frac{q\left(Q-q\right)}{4\pi {\epsilon }_0{r}^2}\Rightarrow \frac{dF}{dq}=\frac{1}{4\pi {\epsilon }_0{r}^2}\left(Q-2q\right)\Rightarrow \frac{dF}{dq}=\frac{1}{2\pi {\epsilon }_0{r}^2}$

Here r is fixed

For maxima or minima of force, its first derivative should be zero.

$\frac{dF}{dq}=\frac{1}{4\pi {\epsilon }_0{r}^2}\left(Q-2q\right)=0\Rightarrow q=\frac{Q}{2}$                

Since second derivative is always negative so maxima will occur at this value of q.

$P_z=V_z{l}_z\Rightarrow 0.5=8l_z\Rightarrow l_z=l_P=\frac{1}{16}Amp$

When zener is connected across a potential divider arranged with maximum potential drop across zener diode, then

$$ $V_P=V-V_z=20-8=12$  volt Potential difference across protective resistance RP

$R_P=\frac{V_P}{I_P}=\frac{12}{\frac{1}{6}}=192Amp.$