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Ede4ka [16]
3 years ago
12

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

. Determine the uncertainty for the slope of the line.
Physics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

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Which of the following is an incorrect statement of one of newtons laws of motion?
weqwewe [10]

Explanation:

The newton's laws of motion are:

First law:

  "A body will remain in its state of rest or of uniform motion along a path unless it is acted upon by an external force. ".

This is popularly called the law of inertia.

Second law:

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Third law:

  "action and reaction forces are equal and opposite in direction".

learn more:

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5 0
3 years ago
A deuteron consists of one proton and one neutron. A deuteron moving horizontally enters a uniform, vertical magnetic field of 0
Helen [10]

Answer:

Explanation:

Let the velocity of deuteron be v then force on it in magnetic field

Bqv , B is magnetic field and q is charge on deuteron . This force will provide centripetal force for circular path so

mv² / r = Bqv   m is mass of deuteron and r is radius of circular path

v = Bqr / m

(.5 x 1.6 x 10⁻¹⁹ x 55.6 x 10⁻² )/ 3.34 x 10⁻²⁷

= 13.31 x 10⁶ m /s

6 0
3 years ago
It's not D
DanielleElmas [232]
A. it provides support
5 0
2 years ago
your friend sit in a sked in the snow. if you apply a force of 120 N to them, they have an acceleration of 1.3 m/s2. what is the
levacccp [35]
Need to know the equation for force
F=MA
F is force
M is mass- we need to know the mass
A is acceleration
use "x" for mass
120 N= 1.3x
divide 1.3 in both side
kg unit for mass
X=92.31 kg 
or 
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Hope this helps
7 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
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