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Relativity and shape of objects

A triangular shape spacecraft flies by an observer at 0.95c. When the ship is measured
by an observer at rest with respect to the ship, the distances x and y are found to be 50m and 25m respectively. What is the shape of the ship as seen by an observer who sees the ship in motion along the direction shown in figure?


Asked Mona

Four persons K, L , M, N are initially at the four corners of a square of side d Problem

Four persons K, L , M, N are initially at the four corners of a square of side ‘d’. Each person now starts moving with a uniform speed ‘v’ in such a way that K always moves directly towards L, L towards M, M towards N and N towards K. After what time will they meet?

Gayathri asked.

Answer: Many students do not understand the real situation initially. Every time the persons are approaching each other and hence they will be moving closer and closer as they continue their walking. Finally they’ll reach the centre of the square to meet each other. I’ve tried to visualize the situation below.

From the diagram, we can make out that the resultant displacement by each when they meet will be d/√2 and the component of velocity of each towards the final point (the centre of the square) is v/√2.

Therefore,the time taken = displacement by the component of velocity in its direction = (d/√2) / (v/√2) = d/v

Numerical problem based on screw gauge

When a screw gauge with a LC of 0.01mm is used to measure the diameter of a

Screw Gauge
Screw Gauge

wire, the reading on the sleeve is found to be 0.5 mm and the reading on the thimble is found to be 27 division i)what is the diameter in cm?

Posted by Swati


Diameter = PSR + CSR x LC = 0.5 +27 x 0.01 = 0.77 mm = 0.077 mm

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