All categories

3 years ago

On a Vernier Caliper, how do you know which mark to use on the very top scale?

Physics

1 answer:

yan [13]3 years ago

5 0

<h2><u>Answer</u></h2>

To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use.

<h2><u>Explanation</u></h2>

A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.

The objects to be measured is place between the jaws of the calipers.

The vernier scale has two scales, the vernier scale and the main scale which is the very top scale. <em>To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use. </em>

You might be interested in

Which four equations can be used to solve for acceleration

emmainna [20.7K]

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as $a = \frac{v-u}{t}$

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as $a = 2*\frac{s-ut}{t^{2} }$

So this is second equation for acceleration.

Then from the third equation of motion, $v^{2}- u^{2} = 2* a *s$

the acceleration equation is determined as $a = \frac{v^{2}-u^{2} }{2s}$

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

8 0

3 years ago

An electrical circuit is powered by a 1.5-volt battery and contains a light bulb producing 5 Ohms of resistance. There is also a

Fed [463]

Answer:

if the resistor is fitted in series with the bulb , then the current flowing will be 0.15 A.....

if the resistor if fitted in parallel with the bulb ,

the current flowing will be 0.6 A

Explanation:

total potential difference = 1.5V

when resistors in series ,

total equivalent resistance is = 5 + 5 = 10 ohm

so current = 1.5 ÷ 10 = 0.15

when resistors in parallel ,

total equivalent resistance is = (5 × 5)÷(5 + 5) =2.5 ohm

so current = 1.5÷2.5 = 0.6 A

8 0

3 years ago

What change will most likely increase the strength of a magnetic field produced by an electromagnet? a. Reduce the number of tur

Anettt [7]

Answer:

Option D

Explanation:

When another battery is added to the circuit, the power supplied through the coil and to the magnet becomes greater leading to stronger magnetic field lines being produced.

7 0

3 years ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

3 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

8 0

3 years ago