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3 years ago

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What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri

ate units. Enter positive value if the direction of the force is in the direction of the initial velocity and negative value if the direction of the force is in the direction opposite to the initial velocity. F =

1 answer:

Yakvenalex [24]3 years ago

5 0

Answer:

The force is $-1.67\times10^{5}\ N$

Explanation:

Given that,

Mass of car = 64 kg

Suppose, a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

$v^2-u^2=2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

$0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}$

$a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}$

$a=-2623.45\ m/s²$

We need to calculate the force

Using formula of force

$F=ma$

$F=64\times(-2623.45)$

$F=-1.67\times10^{5}\ N$

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is $-1.67\times10^{5}\ N$

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You perform an experiment with a long column of air and a tuning fork. The column of air is defined by a very long vertical plas

velikii [3]

Answer:

$\lambda=4L=1.33m$

v=343m/s

Explanation:

We have to take into account the expressions

$f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}$

if we assume that 256Hz is the fundamental frequency we have

$f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m$

and for wavelength

$\lambda=4L=1.33m$

hope this helps!!

6 0

3 years ago

Read 2 more answers

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

Bingel [31]

I'm not exactly sure but I'm thinking that it's the last one. Sorry if I'm wrong

4 0

3 years ago

Read 2 more answers

An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (

vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg

angular speed of the disk = 4 rev/s

arm outstretched = 1 m

centripetal force of the disk in the circular path

F = m ω² r

ω = 4 x 2 x π

ω = 25.13 rad/s

F = m ω² r

F = 2 x 25.13² x 1

F = 1263.03 N

hence, centripetal force equal to the F = 1263.03 N

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3 years ago

American football uses a field that is 100.0yd long, whereas the a soccer field is 100.0m long. Which field is longer and by how

Vitek1552 [10]

The soccer field is 9.36% longer.

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3 years ago