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3 years ago

How does uncertainty relate to measurements and calculations made during an investigation?

Physics

2 answers:

Nana76 [90]3 years ago

6 0

Answer:

Explanation :Uncertainty in measurements and calculations means difference between actual and measured data. We can say that all measurements have some degree of uncertainty. It can be of two types-1. Systematic error (because of error in measuring instrument) 2. Random error (human errors such as- delay in starting, delay in stopping).

Systematic errors are consistent in magnitude and direction so they can be easily removed by additive or proportional corrections while random errors vary in magnitude and direction and can not be easily removed.

DerKrebs [107]3 years ago

4 0

Uncertainty means that your result may be very random, so you can't trust the first or second or so observation, making several samples critical for accuracy.

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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.

Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0

3 years ago

PLZ HELP I WILL GIVE BRAINLIEST

Vaselesa [24]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of diver = 77kg

Height of jump = 8.18m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we apply the motion equation below:

v² = u² + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

Now insert the parameters and solve;

v² = 0² + 2 x 9.8 x 8.18

v = 12.7m/s

8 0

2 years ago

Help please asap I'm super lost

vaieri [72.5K]

A compound is two or more elements combinded together.
A element is a pure substance.
it looks like all of them are compounds except the carbon dixode, My teacher would accept that as a anwser idk bout urs thoughs.

6 0

3 years ago

Read 2 more answers

A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip

Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

Given the following data:

Height = 1.45 meters

Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the height or distance covered.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

$1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0$

The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}$

Substituting the values, we have;

$t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}$

Time, t = 0.151 seconds.

Therefore, it will take the quarter 0.151 seconds to reach the ground.