Can someone please explain some calculations on vernier calliper

2 answers:

6 0

The Vernier caliper is an instrument that allows you measure lengths much more accurate than the metric ruler. The smallest increment in the vernier caliper you will be using is (1/50)mm = 0.02mm = 0.002cm. Thus, the uncertainty is ∆x = (1/2)0.002 cm = 0.001 cm. Vernier caliper least counts formula is calculated by dividing the smallest reading of the main scale with the total number of divisions of the vernier scale.LC of vernier caliper is the difference between one smallest reading of the main scale and one smallest reading of vernier scale which is 0.1 mm 0r 0.01 cm.

laiz [17]3 years ago

3 0

Answer:

Hey mate here's your answer ⤵️

Vernier caliper least counts formula is calculated by dividing the smallest reading of the main scale with the total number of divisions of the vernier scale.LC of vernier caliper is the difference between one smallest reading of the main scale and one smallest reading of vernier scale which is 0.1 mm 0r 0.01 cm

<h2>Hope it was helpfulll</h2>

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A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are

gavmur [86]

1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

$F = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 275 N

Therefore, the coefficient of static friction is

$\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935$

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

$F = \mu_k mg$