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2 years ago

A body of mass 11 kg is subjected to a net force of 20 N East for 30 s. Calculate the change in

Physics

1 answer:

Vanyuwa [196]2 years ago

5 0

Answer:

54.55 m/s

Explanation:

From the question above,

Change in velocity = impulse/mass

Δv = I/m = Ft/m................... Equation 1

Where F = force, t = time, m = mass

Given: F = 20 N, t = 30 s, m = 11 kg

Substitute these values into equation 1

Δv = (20×30)/11

Δv = 600/11

Δv = 54.55 m/s

Hence the change in velocity of the body is 54.55 m/s

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Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

consider the attached diagram below

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )$

$V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )$

Answer

$V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }$

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

8 0

2 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

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2 years ago

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