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Ede4ka [16]
4 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]4 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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A hungry 177177 kg lion running northward at 75.575.5 km/hr attacks and holds onto a 32.332.3 kg Thomson's gazelle running eastw
ValentinkaMS [17]

Answer: 73.2 km/hr

Explanation:

This is a momentum conservation type of problem, here we need to use the fact that the initial momentum is equal to the final momentum.

Remember that the momentum is written as:

P = m*v

m = mass

v = velicity

At the beginning, we have:

A lion of 177 kg running at velocity of 75.5 km/hr

Then the momentum of the lion is P1 = 177 kg*75.5 km/hr= (13,363.5)kg*km/hr

The gazelle has a mass of 32.3 kg, and a velocity of 60.6 km/hr

Then the momentum of the gazelle is:

P2 = 32.3kg*60.6km/hr = 1,957.38 kg*km/hr

The total initial momentum is equal to P1 + P2, then we have:

P = 1,957.38 kg*km/hr + 13,363.5 kg*km/hr = 15,320.9 kg*km/hr

Now, after the lion catches the gazelle, we can think it as only one object moving with mass equal to the mass of the lion plus the mass of the gazelle and velocity V

Then the final momentum will be:

Pf = (177kg + 32.3kg)*V = 209.3kg*V

And this needs to be equal to the initial momentum, then:

209.3kg*V = 15,320.9 kg*km/hr

V = (15,320.9 kg*km/hr)/209.3kg = 73.2 km/hr

Then the final speed of the lion-gazelle system is 73.2 km/hr

4 0
3 years ago
Which of the following does not dissolve in water​
Anna11 [10]

Answer:

Sand,Mud

Explanation:

Sugar and salt are examples of soluble substances. Substances that do not dissolve in water are called insoluble. Sand and flour are examples of insoluble substances.

6 0
3 years ago
Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius R = 7.5 cm. Assume that the powder
Ipatiy [6.2K]

Answer:

A) E =\frac{\rho r}{2\epsilon}

B) v = 58.7923\times 10^4 V

Explanation:

a) using Guass law

\oint E.dA = \frac{q_{enclosed}}{\epsilon_o}

EA = \frac{q_{enclosed}}{\epsilon_o}

E =  \frac{q_{enclosed}}{A \epsilon_o}

HereA is = 2\pi rL

E =  \frac{q_{enclosed}}{(2\pi rL) \epsilon_o}

Volume charge density is given as

\rho = \frac{q_{enclosed}}{volume}

net charge is given as

q_{enclosed} = \rho \times volume

therefore E =  \frac{ \rho \times (L\pi r^2)}{(2\pi rL) \epsilon_o}

VOLUME  =  L\pi r^2

After solving electric field equation we get

E =\frac{\rho r}{2\epsilon}

b) electric potential difference is given as

v_{wall} - v = - \int_{r}^{R} Edr

0 - v = - \int_{r}^{R} E dr

v = \int_{r}^{R} Edr

= \int_{r}^{R} (\frac{\rho r}{2\epsilon}) dr

= \frac{\rho}{4\epsilon} (R^2 - r^2)

at r = 0

v = \frac{- 3.7 \times 10^{-3} \times 0.075^2}{4\times (8.85\times 10^{-12}}

v = 58.7923\times 10^4 V

7 0
3 years ago
What is the unit for electric charge?<br> Kilogram<br> Newton<br> Electron<br> Coulomb
Valentin [98]

Answer:

Coulomb.

Explanation:

6 0
3 years ago
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An object has a mass of 60 kg. It decelerates from 50 m/s to 20 m/s when a resultant force of 300 N acts on it. For how long doe
Pavlova-9 [17]
  • Force=300N
  • Mass=60kg

Find Acceleration

\\ \rm\longmapsto F=ma

\\ \rm\longmapsto a=\dfrac{F}{m}

\\ \rm\longmapsto a=\dfrac{-300}{60}

\\ \rm\longmapsto a=-5m/s^2

Now

  • u=50m/s
  • v=20m/s

Using 1st equation of kinematics

\\ \rm\longmapsto v=u+at

\\ \rm\longmapsto t=\dfrac{v-u}{a}

\\ \rm\longmapsto t=\dfrac{20-50}{-5}

\\ \rm\longmapsto t=\dfrac{-30}{-5}

\\ \rm\longmapsto t=6

4 0
3 years ago
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