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3 years ago

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A battery having an emf of 9.76 V delivers 111 mA when connected to a 60.5 Ω load. Determine the internal resistance of the batt

ery. ______Ω

1 answer:

kow [346]3 years ago

5 0

Answer:

$r = 27.4 \: ohm$

Explanation:

Using

E = I(R + r)

$9.76 = 111 \times {10}^{ - 3} (60.5 + r) \\ 87.928 = 60.5 + r \\ r = 27.4 \: ohm$

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

Radius, $r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

Velocity, $v=\frac{Bqr}{m}$

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

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3 years ago

One negative change you may encounter as a student or an employee

matrenka [14]

Answer:

you may get bullied or teased for being a differrent race, ethnic.

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Light moves at a speed of around 1 million miles per hour<br>O<br>True<br>False

AURORKA [14]

Answer:

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Explanation:

In miles per hour, light speed is about 670,616,629 mph

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An object kept in front of lens, which is thick at the edges, produces a _________,

Wewaii [24]

Answer:

B

Explanation:

A diverging lens forms a virtual image that is as the same size as the object

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3 years ago

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

Svetradugi [14.3K]

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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