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2 years ago

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Kinetic energy can be calculated using the formula KE = 1/2mv2. What is the kinetic energy of a bicycle that has a mass of 20 kg

and is traveling at a velocity of 10 m/sec?

1 answer:

Alenkasestr [34]2 years ago

8 0

<h3>Answer:</h3>

KE = 1000 J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Physics</u>

<u>Energy</u>

Kinetic Energy Formula: $\displaystyle KE = \frac{1}{2}mv^2$

Energy is in Joules
m is mass (in kg)
v is velocity (in m/s)

<h3>Explanation:</h3>

<u>Step 1: Define</u>

m = 20 kg

v = 10 m/s

<u>Step 2: Find KE</u>

Substitute [KE]: $\displaystyle KE = \frac{1}{2}(20)(10)^2$
Exponents: $\displaystyle KE = \frac{1}{2}(20)(100)$
Multiply: $\displaystyle KE = (10)(100)$
Multiply: $\displaystyle KE = 1000$

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

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Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

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making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

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