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3 years ago

What do you meant by varnier constant?

Physics

1 answer:

Soloha48 [4]3 years ago

3 0

Answer:

the ratio of the smallest division of main scale to the number of divisions of the vernier scale.

Explanation:

difference between the value of one main scale division and one vernier scale division

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A car covers 120 km in 3 hours calculate its speed in m/s

8_murik_8 [283]

Answer:

11.1 m/s

Explanation:

120 km = 120 x 1,000 = 120,000 m

3 hours = 3 x 60 x 60 = 3600 x 3 = 10,800 s

speed = 120,000 / 10,800

= 1200/108

= 11.1 m/s

6 0

2 years ago

Read 2 more answers

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

3 years ago

Find the uniform acceleration that causes a car's velocity to change from 20.0 m/s to 105 m/s in and 12.0 s

Sauron [17]

Answer:

7.08 m/s²

Explanation:

Given:

v₀ = 20.0 m/s

v = 105 m/s

t = 12.0 s

Find: a

v = at + v₀

105 m/s = a (12.0 s) + 20.0 m/s

a = 7.08 m/s²

5 0

3 years ago

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

ololo11 [35]

Answer:

Part a)

$f_B = 290 Hz$

Part B)

percentage increase is

$percentage = 1.38$ %

Explanation:

Part a)

As we know that the beat frequency is

$f_A - f_B = 3$

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

$293 - f_B = 3$

$f_B = 290 Hz$

Part B)

percentage increase in the tension of the string will be given as

$f_A - f_B' = 1$

$f_B' = 292 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

so we have

$T_1 = C (290)^2$

$T_2 = C(292)^2$

so we have

$\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}$

percentage increase is

$percentage = 1.38$

4 0

4 years ago

Cual sera el caudal que lleva un rio cuando se desplaza a 200 litros cada 40 segundos

alisha [4.7K]

Answer:

Q = 5 L/s

Explanation:

To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

$Q=\frac{V}{t}$

V: Volume (volumen) = 200L

t: time (tiempo) = 40 s

you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

$Q=\frac{200L}{40s}=5\frac{L}{s}$

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)

4 0

4 years ago