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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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At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

6 0

3 years ago

800 joules of work were done with a force of 200 newtons. Over what

Svetach [21]

Answer:

Explanation:

800/200

7 0

3 years ago

Neglecting air resistance, the distance s left parenthesis t right parenthesis in feet traveled by a freely falling object is g

ivann1987 [24]

Explanation:

It is given that, the height of a certain tower is 862 feet i.e to reach on the ground the object should travel, s = 862 feet

The distance traveled by a freely falling object is given by :

$s(t)=16t^2$

$16t^2=862$

$t=\sqrt{53.875}$

t = 7.34 seconds

So, the object will take 7.34 seconds to fall to the ground from the top of the building. Hence, this is the required solution.

7 0

3 years ago

Una bola de acero rueda y cae por el borde de una mesa desde 4ft por encima del piso. Si golpea el suelo a 5ft de la base de la

ycow [4]

Answer:

$\large\boxed{\large\boxed{10ft/s}}$

Explanation:

The question, translated, is:

A steel ball rolls and falls off the edge of a table from 4ft above the floor. If you hit the ground 5ft from the base of the table, what was your initial horizontal velocity?

<h2>Solution</h2>

This is a projectile motion, for which, the equations that you will need are:

$V_x_0= V_x=x\cdot t$

$y=y_0+Vy_o\cdot t-g\cdot t^2/2$

1. Calculate the time that it takes the ball to fall 4ft

$0=4ft-g\cdot t^2/2\\\\t^2=2\times 4ft/( 32.174ft/s^2)=0.24865s^2\\\\t=0.4986s$

2. Calculate the horizontal velocity:

$V_x_0= V_x=x\cdot t\\\\V_x_0=5ft/0.4986s=10.027ft/s\approx 10ft/s$

3 0

3 years ago

A marble rolling at a speed of 2.71 m/s falls of the end of a table that is 1.25 m high. How far from the base of the table does

katrin [286]

Answer:

The marble lands at a distance of 1.36 m from the base of the table.

Explanation:

The motion of the marble falling off the table is a projectile with initial velocity in the horizontal direction only.

The motion can be solved in two directions, the horizontal and vertical direction.

Along the vertical direction, the initial velocity is 0 m/s as it has only horizontal component initially. Also acceleration in the vertical direction is acceleration due to gravity.

Let us use equation of motion in vertical direction.

$y-y_0=v_{0y}t+\frac{1}{2}at^2$

Where,

$v_{0y}\rightarrow \textrm{vertical component of the initial velocity}.\\y\rightarrow \textrm{final position of the marble}\\y_0\rightarrow \textrm{initial vertical position}\\a_y\rightarrow \textrm{acceleration in the vertical direction}\\t\rightarrow \textrm{time taken to reach bottom}$

Now, plug in 0 for $y$ , $1.25$ for $y_0$ , 0 for $v_{0y}$ , -9.8 for $a_y$ . This gives,

$0-1.25=0+\frac{1}{2}(-9.8)t^2\\-1.25=-4.9t^2\\t^2=\frac{-1.25}{-4.9}\\t^2=0.255\\t=\sqrt{0.255}=0.501\ s$

Therefore, time to reach bottom is 0.501 s.

Now, consider the horizontal motion. There is no acceleration in the horizontal direction. So, distance is given as the product of horizontal velocity and time taken.

Horizontal distance covered by the marble is given as:

$x=v_{0x}\times t=2.71\times 0.501=1.36 \textrm{ m}$

Therefore, the marble lands at a distance of 1.36 m from the base of the table.

6 0

3 years ago