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Mrac [35]
3 years ago
15

You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from

walking on carpeting you have accumulated a net negative charge. Your car is considered a neutral conductor. As you place your hand near the door handle (without touching it) what happens to the charges of your car?
Physics
1 answer:
xxMikexx [17]3 years ago
4 0

Answer:

There is a localization of negative charge near the door handle.

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A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

                    = 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A)  angular frequency = ?

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k  x - ω t)

y ( x ,t) = 2.34  x  sin(33.59 x - 12.75 t)

4 0
3 years ago
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
Which of these events is associated with September 11, 2001?
Oxana [17]
Terrorist attacks on the United States is the answer.

On September 11, 2001 that was the day New York got attacked by terrorists. The Twin Towers were the ones that got affected, 2,996 people (maybe more) died during that attacked. The terrorists were one of the 2,996 people that died (19 of them died). more than 6,000 were injured that day.
#NeverForget

Hope this helped
Have a great day<span />
7 0
3 years ago
A tightrope walker more easily balances on a tightwire if his pole
cestrela7 [59]
B) droops.

Why?
To maintain balance, you do not need something short so you're balanced well... You need something long and droopy to maintain balance. The pole should be held by your waist and it should be light.

Hope this helps!~
4 0
3 years ago
It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?
ankoles [38]
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed 
 Vf = 27.5 m/s  Final speed 

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2  -24.0 ^2)
160kj =  4680 x m
convert kilo joules to jeoules                     160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
4 0
3 years ago
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