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zloy xaker [14]
3 years ago
9

The sum of the digits of a two-digit number is 6. When the digits are reversed, the number decreases by 18. Find the original nu

mber.
Physics
1 answer:
Setler [38]3 years ago
7 0
<h2>The number is 42     </h2>

Explanation:

Let the digits be a and b.

The sum of the digits of a two-digit number is 6

That is

         a + b = 6

When the digits are reversed, the number decreases by 18

             ab - ba = 18

We need to find the number

Since ab - ba = 18, we know that a is greater than b.

Take a = 4

                4 + b = 6

                      b = 6 - 4 = 2

 That is the number is ab = 42

  When it is reversed the number is 24.

              42 - 24 = 18, hence correct

The number is 42          

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

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u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
4 years ago
Two balls have masses 18 kg and 47 kg. The 18 kg ball has an initial velocity of 76 m/s (to the right) along a line joining the
BigorU [14]

Answer:

The final velocity of 18 kg ball is V_{2} = 42.09 \frac{m}{sec}

Explanation:

Mass of first ball m_{1} = 18 kg

Mass of second ball m_{2} = 47 kg

Initial velocity of 18 kg ball V_{1} = 76 \frac{m}{sec}

Initial velocity of 47 kg ball = 0

Final velocity of 18 kg ball V_{2} = ??

Final velocity of 18 kg ball  is given by the formula

V_{2} = \frac{2 m_{1} V_{1} }{m_{1} + m_{2}  }

Put all the values in above formula we get

V_{2} = 2 × 18 × \frac{76}{65}

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3 0
3 years ago
Question 3 of 10
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Answer:

30N

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The mechanical energy of a 2kg body is 35J and its potential energy is 10J calculate speed energy​
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5 0
3 years ago
A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r
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Answer:

0.0109 m ≈ 10.9 mm

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<u>determine the radius of the circle in which an electron would move </u>

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For proton :

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For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

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= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0
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