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2 years ago

How can you find the reading of main scale and vernier scale

Physics

1 answer:

anygoal [31]2 years ago

7 0

Answer:

this pdf should help you out

Explanation:

Download pdf

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A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v

IRISSAK [1]

The solution to this task is f00king Dis.

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3 years ago

Find the acceleration produced.

krok68 [10]

Answer:

0.67m/s²

Explanation:

Given parameters:

Mass of toy = 1.2kg

Force applied = 0.8N

Unknown:

Acceleration = ?

Solution:

According to newton's second law of motion;

Force = mass x acceleration

Now,

Acceleration = $\frac{Force}{mass}$

Acceleration = $\frac{0.8}{1.2}$ = 0.67m/s²

4 0

3 years ago

Read 2 more answers

The latitude of any location on earth is the angle formed by the two rays drawn from the center of earth to the location and to

Alina [70]

The distance between city a and city b is 833.345 miles.

We know that

1°=60'

The distance of city a from the initial ray is calculated as

x_a=3960*tan45.46°=4024.101 miles

The distance of city b from the initial ray is calculated as

x_b=3960*tan 38.86°=3190.75 miles

Now the distance between city a and b is equal to

4024.101-3190.75=833.345 miles

This is the vertical distance between the cities.

5 0

3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t

navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that $v^2=u^2+2as$

$16^2=24^2+2\times a\times 50$

$a=-3.2m/sec^2$

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So $v^2=u^2+2as$

$0^2=24^2-2\times 3.2\times s$

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration $a=-3.2m/sec^2$

So time $t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec$

7 0

3 years ago

How can you find the reading of main scale and vernier scale​

How can you find the reading of main scale and vernier scale