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3 years ago

Difference between reproducibility and replicability

1 answer:

8 0

Suppose you are doing an experiment where you determine the value of one parameter, say density of a liquid. You have two methods in doing this. By finding the mass and volume, and by using a densitometer. Reproducibility is when you get the same value of density for both methods. Replicability is when you have similar results in one method. So, replicability is a measure of precision, while reproducibility is a measure of accuracy.

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5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0

3 years ago

A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running

Katena32 [7]

Answer:

$t_a = 13.49 \ s$

The distance is $D = 55.2 \ m$

Explanation:

From the question we are told that

The maximum speed of the cheetah is $v = 103 \ km/h = 28.61 \ m/s$

The maximum of gazelle is $u = 76.5 \ km/h = 21.25 \ m/s$

The distance ahead is $d = 99.3 \ m$

Let $t_a$ denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

$v* t_a = d + u* t_a$

=> $28.61 t_a = 99.3 + 21.25t_a$

=> $7.36 t_a = 99.3$

=> $t_a = 13.49 \ s$

Now at t = 7.5 s

$7.5 v = D+ 7.5u$

=> $28.61 * 7.5 = D + 21.25* 7.5$

=> $7.36 * 7.5 =D$

=> $D = 55.2 \ m$

Hence the for the gazelle to escape the cheetah it must be 55.2 m

5 0

3 years ago

The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ahrayia [7]

<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:

1. The force of air friction, also called "drag force" $D$ :

$D={C}_{d}\frac{\rho V^{2} }{2}A$ (1)

Where:

$C_ {d}$ is the drag coefficient

$\rho$ is the density of the fluid (air for example)

$V$ is the velocity

$A$ is the transversal area of the object

So, this force is proportional to the transversal area of the falling element and to the square of the velocity.

2. Its weight due to the gravity force $W$ :

$W=m.g$

(2)

Where:

$m$ is the mass of the object

$g$ is the acceleration due gravity

So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity:

$D=W$ (3)

${C}_{d}\frac{\rho V^{2} }{2}A=m.g$ (4)

$V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}$ (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0

2 years ago

Which of the following statements is true?

hram777 [196]

All the radiation from stars is the result of nuclear fusion in their cores.

4 0

3 years ago

Read 2 more answers

An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.

Marianna [84]

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

Given the following data;

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

4 0

3 years ago