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3 years ago

Why are both accuracy and reproducibility important in making a measurement?

1 answer:

4 0

Accuracy is how close you're measurement comes to an accepted or given value. I n many cases you do not know what the accepted value is, so you have nothing to compare your measurement with. The more often a measurement is taken with close precision, or reproducibility, the more likely you are to being close to your unknown accepted value.
There is a great short tutorial video covering accuracy and precision at Sciocity dot com

You might be interested in

If for every action force an equal and opposite reaction force exists, how can anything ever be accelerated?

sweet [91]

If there are two equal and opposite forces on the SAME THING, then the thing doesn't accelerate. You're right about that. But the action and reaction forces act on two different things. The bullet and the rifle. The ball and the bat. The airplane and the air. etc. So BOTH can accelerate.

8 0

3 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

2 years ago

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

3 years ago

The vibration produced days or even years before an earthquake

Margaret [11]

I don't know I guess its the plate tectonics

7 0

2 years ago