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zlopas [31]
3 years ago
6

An electron is carried from the positive terminal to the negative terminal of a 9 v battery. How much work is required in carryi

ng this electron?
Physics
1 answer:
shutvik [7]3 years ago
5 0

Answer:

The work required for the electron is 1.44x10^{-18} J

Explanation:

The work of the electron is calculated in the equation:

W= e * V

e= 1.6x10^{-19} C

C= Coulomb = s*A

V=  Volts = \frac{W}{A}

W= \frac{J}{s}

W= 1.6 x10^{-19}C *9 V\\W=1.44x10^{-18} C*V\\ W=1.44x10^{-18} s*A*\frac{W}{A} \\W=1.44x10^{-18} W*s\\W=1.44x10^{-18} \frac{J}{s}*s \\W=1.44x10^{-18} J\\

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dimulka [17.4K]

Answer:

Velocity is 1.73 m/s along 54.65° south of east.

Explanation:

Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 2i + m x (-1)i = m i

Final momentum = m x v + m x 1.41 j = mv + 1.41 m j

Comparing

mi = mv + 1.41 m j

v = i - 1.41 j

Magnitude of velocity

      v=\sqrt{1^2+(-1.41)^2}=1.73m/s        

Direction,  

       \theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0             

Velocity is 1.73 m/s along 54.65° south of east.

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3 years ago
A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb
Nostrana [21]

Explanation:

It is given that,

Velocity in East, v_1=4\ m/s

Velocity in North, v_2=3\ m/s

(a) The resultant velocity is given by :

v=\sqrt{4^2+3^2}=5\ m/s

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

t=\dfrac{d}{v}

t=\dfrac{80\ m}{5\ m/s}

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

x=v_2\times t

x=3\ m/s\times 16\ s

x = 48 meters

Hence, this is the required solution.

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3 years ago
What would happen to the coil if the current passing through it was too large?
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What is an atom? what are electrons,protons,and neutrons?
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2 years ago
Read 2 more answers
Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 32 ∘ with the vertical.
shtirl [24]

To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.

From the image:

cos (32) = \frac{(7.6-h)}{7.6}

h = 1.1548m

When Jeff reaches his lowest point his potential energy is converted to kinetic energy

PE = KE

mgh = \frac{1}{2} mv^2

v = \sqrt{2gh}

v = \sqrt{2(9.8)(1.1548)}

v= 4.75m/s

Tension in the string at the lowest point is sum of weight of Jeff and the his centripetal force

T = W+F_c

T = mg + \frac{mv^2}{r}

T = (83)(9.8)+\frac{(9.8)( 4.75)^2}{7.6}

T = 842.49N

Therefore the tension in the vine at the lowest point of the swing is 842.49N

3 0
2 years ago
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