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2 years ago

7

YALL THIS QUESTION IS DUE IN 2 HOURS AND I HAVE 3 MORE PAGED TO DO PLEASE HELP ME WITH THIS PHYSICS HOMEWORK PROBLEM!!!! I ATTAC

HED A PIC OF THE PROBLEM BELOW!!!!

1 answer:

goldfiish [28.3K]2 years ago

7 0

Answer:

I dont know :)

Explanation:

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What is the acceleration of the object?<br> m/s2

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Answer:use

Explanation:google

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3 years ago

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object $m=2\ kg$

Speed of object $u=5\ m/s$

Mass of object at rest $M=3\ kg$

Suppose after collision, speed is v

conserving momentum

$\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s$

Initial kinetic energy

$k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J$

Final kinetic energy

$k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J$

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

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2 years ago

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

<em>consider the attached diagram below</em>

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

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

OR

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

8 0

3 years ago

You carry a 5 kg sack of potatoes across the store for 5 minutes, and you wait in line holding it for 30 seconds before you get

ludmilkaskok [199]

Answer:

The only work done is when the person lifts the sack over a distance, W = 78.48 [N]

Explanation:

We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.

W = F * d

where:

F = force [N]

d = distance [m]

The force is given by the producto of the mass by the gravity.

F = 5 * 9.81 = 49.05 [N]

W = 49.05 * 1.6 = 78.48 [N]

3 0

3 years ago

One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that

Rufina [12.5K]

Answer:

45930.52N

Explanation:

Net force = (internal pressure - external pressure)× area of window

Net force = (1.02 - 0.910)atm × 2.03m × 2.03m = 0.11atm × 4.1209m^2 = 0.11 × 101325N/m^2 × 4.1209m^2 = 45930.52N

5 0

3 years ago