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ziro4ka [17]
3 years ago
9

Doing a physics Lab and need to propagate uncertainty for experimental results. I need the uncertainty in order to add it to my

scatter plot for the error bars, therefore, I need both a positive and a negative error value. How is this done for my data? please show all work
373.67 +/- 15.444

210 +/- 12.82

239.33 +/- 19.98

443.67 +/- 17.99
Physics
1 answer:
alexandr1967 [171]3 years ago
8 0

Answer:

Explanation:

Not really sure what you're trying to do. You propagate uncertainties for indirect measurements, as in when you calculate a value from other values.

What you have here is a series of values of direct measurements it seems.

Anyway, for error bars will have a width of 2 times the uncertainty reported.

For example on the first one

373.67 +/- 15.444

You would have an error bar with a width of 2 * 15.444 = 30.888. This bar would be centered at 373.677. The lowest point of the error bar would be at 358.233 and the highest point at 389.121.

You also mentioned a scatter plot, but scatter plots are 2D at least. Are these measurements associated to something else like time? You need 2 coordinates for each point in a scatter plot.

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Because the fossil show layers and those layers show what kind of settlements where left behind at that layer
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You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
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a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

from above statement we got

height = 78.4 m

since the ball is thrown, so its vertical velocity would be zero

u = 0

taking g = 9.8m/s^2

now, using the equation of motion

h = ut + gt^2/2

now putting all the values in it

we got ,

78.4 = 9.8 * t^2/ 2

by solving we got,

t = 4 sec

b) now, since along the horizontal , no force acting and accelaration is zero so

R = ut , R is RANGE

R = 5 * 4

range =  20 m

c)  vertical components of the stone’s velocity just before it hits the ground = v sin θ =

horizontal   components of the stone’s velocity  just before it hits the ground = v cos θ

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4 0
1 year ago
Unit 5 activity 2 I need answer
saul85 [17]

Answer:

Can you explain more in detail what it is?

Explanation:

6 0
3 years ago
Nuclear power is regarded as a clean source of energy because it doesn't produce or
Nadya [2.5K]

Emit greenhouse gases

Explanation:

Nuclear power is regarded as a clean energy source because it doesn't produce or emit greenhouse gases.

Greenhouse gases are mostly generated by the use of fossil fuels that pollutes the environment causing surging global temperature.

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There is no production of any greenhouse gas in the process of harnessing nuclear energy.

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8 0
4 years ago
A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg an
Ainat [17]

Answer:

Explanation:

Momentum is the product of mass of a body and its velocity.

Given the mass of the puck m1 = 0.16kg

velocity of the puck v1 = 40m/s

Given the mass of the goalkeeper m2 = 120kg

velocity of the goalkeeper v2= 0m/s (goal keeper at rest)

The total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper is expressed as:

m1v1 + m2v2 (their momentum will be added since they collide)

= 0.16(40) + 120(0)

= 0.16(40) + 0

= 6.4kgm/s

Let us calculate their common velocity using the conservation of momentum formula;

m1u1 + m2u2 = (m1+m2)v

6.4 = (0.16+120)v

6.4 = 120.16v

v = 6.4/120.16

v = 0.053m/s

Hence after collision, both objects move at a velocity of 0.053m/s

Momentum of the puck after collision = m1v

Momentum of the puck after collision = 0.16*0.053m/s

Momentum of the puck after collision = 0.0085kgm/s

Momentum of the keeper after collision = m2v

Momentum of the keeper after collision = 120*0.053m/s

Momentum of the keeper after collision = 6.36kgm/s

From the calculation above, it can be seen that the keeper has the greater momentum after the puck was caught since the momentum of the keeper after collision is greater than that of the puck

4 0
3 years ago
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