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Elza [17]
3 years ago
7

If the reading of a linear scale is 4 mm and no of division of the circular scale is 50, then what will be the diameter of the w

ire in mm? (Least count = 0.01) * 2 points 2.25 3.5 4.5 9.0
Physics
1 answer:
Sliva [168]3 years ago
8 0

Answer:

Diameter of Wire = 4.5 mm

Explanation:

First, we need to find the fractional part of the reading. The fractional part o the reading can be given by the following formula:

Fractional Part = Circular Scale Reading x Least Count

where,

Circular Scale Reading = 50

Least Count = 0.01 mm

Therefore,

Fractional Part = (50)(0.01 mm)

Fractional Part = 0.5 mm

Now, the diameter of the wire can be given by using the following formula:

Diameter of Wire = Linear Scale Reading + Fractional Part

Diameter of Wire = 4 mm + 0.5 mm

<u>Diameter of Wire = 4.5 mm</u>

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Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi
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Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

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Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

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