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3 years ago

Why is it important to consider the experimental error in all the empirical results presented?

Physics

2 answers:

irinina [24]3 years ago

6 0

Answer:

It is very important because scientists, especially the ones with empirical experiments and results, are prone to error and the empirical data is in need to be under strict observation done not only by many scientists but also by expermiented ones. This guards everybody to change the parameters suddenly which can affect the real results of an experiment

Explanation:

brilliants [131]3 years ago

3 0

Answer:

Explanation:

Suppose you want to know how many miles a tire gets before it must be replaced.

Suppose further that you live in a place where the roads are not very good. You would be quite concerned about how you were going to take road conditions into account.

Is rain different from snow? Will the tire get more miles in rain or snow?

Does the tire respond differently after the rain has stopped? It likely will, but how differently.

What about getting stuck especially if an alligator is the only thing looking at you being stuck and he has lunch written all over his face. Is the tire able to get you out?

All of these factors will be very important because no two seasons are alike and one July (say) of one year is not the same as another July in a different year. There has to be a way of determining what the numbers mean.

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A 930-kg sports car collides into the rear end of a 2000-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

bekas [8.4K]

Answer:2.103 m/s

Explanation:

Given

mass of sports car $m=930\ kg$

mass of SUV $M=2000\ kg$

Suppose u is the velocity if sports car before collision

Conserving momentum we get

$mu=(M+m)v$

$v=\dfrac{2000+930}{930}\times u$

$v=3.15\cdot u$

After collision the combined mass drag 2.8 m and finally stops

From work energy theorem work done by friction is equal to change in kinetic energy of the combined mass system

$\frac{1}{2}(M+m)v^2=\mu (M+m)gx$

where $\mu =\text{coefficient of friction}$

$x=\text{drag distance}$

$v=\sqrt{2\mu gx}$

$v=\sqrt{2\times 0.8\times 9.8\times 2.8}$

$v=6.626\ m/s$

Initial velocity $u=\frac{6.626}{3.15}$

$u=2.103\ m/s$

6 0

3 years ago

A. What is the wavelength?

o-na [289]

Answer:

Wavelength is the distance between successive crests of wave, especially points in a sound wave or electromagnetic

wave.

Explanation:

<h3>I hope it helps ❤❤</h3>

4 0

3 years ago

Read 2 more answers

Let’s take the case of the 4-kg block with an initial velocity of 10 m/s that is colliding with a 6-kg block that is stationary.

Aleksandr [31]

Answer:

6kg ise yes...............

8 0

2 years ago

Read 2 more answers

A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha

Vlada [557]

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
$\Delta U = q \Delta V = q (V_f - V_i)$
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: $q=1.6 \cdot 10^{-19} C$ , and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
$\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J$

7 0

3 years ago

Read 2 more answers

What is the net force on a 4,000-kg car

aniked [119]

Answer:

6000 N toward west

Explanation:

F = ma

a = 30-15/10 = 1.5 m/s^2 towards west

m = 4000 kg

F = 4000 x 1.5 = 6000 N towards west

8 0

3 years ago