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3 years ago

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

Physics

1 answer:

Rom4ik [11]3 years ago

3 0

there here is your answer to your question

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

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A tissue is made of a group of similar what working together

matrenka [14]

Cells.

When you have a group of similar cells, you make tissues. This is like having many specialized muscle cells grouping together to make a muscle tissue that can then make something like a bicep.

3 0

3 years ago

The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

Stels [109]

Answer:

$T_1=0.24y$

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

$(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3$

Where $T_1$ and $T_2$ are the orbital periods of Mercury and Earth respectively. We have $D_1=0.39D_2$ and $T_2=1y$ . Replacing this and solving for

$T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y$

5 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

4 years ago

Why is rock away from ocean ridges older? Is it because if the rocks are away from cracks in the crust, there is no way for any

andreyandreev [35.5K]

Answer:

the rocks are away from cracks in the crust

Explanation:

The new crust is pushed away from the Ocean Ridge in both directions as newer crust is formed. This is called sea floor spreading. The crust that makes up the sea floor starts to have time to accumulate a layer of sediments as it gets older and moves away from the Ocean Ridge.

6 0

2 years ago

Read 2 more answers

HELP

sergiy2304 [10]

Answer:

Explanation:

a charged and uncharged object attratct eachother that is the answer your welcome

8 0

3 years ago