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3 years ago

What is The Difference between Accuracy and Precision?

2 answers:

7 0

In other words, accuracy describes the difference between the measurement and the part's actual value, while precision describes the variation you see when you measure the same part repeatedly with the same device

GrogVix [38]3 years ago

7 0

Answer:

Accuracy describes the difference between measurement and the parts actually value, while precision describes the variation you see when you measure the same part repeatedly with the same device.

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Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608\times 10^{-9}\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=\dfrac{mD\lambda}{a}$

where

a = width of the slit

$a=\dfrac{mD\lambda}{y}$

$a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}$

a = 0.000167 m

$a=1.67\times 10^{-4}\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

7 0

3 years ago

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

$\frac{v_o}{r_o} = \frac{v_1}{r_1}$

v1 = $\frac{r_1}{r_o} \ \ v_o$

let's calculate

v₁ = $\frac{1.50}{1.00} \ \ 25.0$

v₁ = 37.5 cm / s

4 0

3 years ago

at the extact same time, youj drop a piece of notebook paper and your science out the window. the science book hit first? why?

jeyben [28]

The piece of paper has less mass and will glide down the window, whereas the textbook will go straight to the ground. Since the textbook has more mass and less ways of it being able to 'glide' the textbook will hit the ground first.

8 0

3 years ago

A circular loop of wire lies flat on a level table top in a region where the magnetic field vector points straight upward. The m

jarptica [38.1K]

The magnetic field direction and direction of induced current in a wire are related by the right hand grip rule. Since the magnetic field was upwards, the thumb points upwards and the fingers curl around it. When viewed from above, it is seen as a current flowing in the counter clockwise direction.

4 0

3 years ago

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h

zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s

Induced peak emf = NABω

= 500 x (π x 0.25²) x 0.425 x 376.738

= 15721.16 V

= 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0

3 years ago