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leva [86]
2 years ago
9

A housefly walking across a surface may develop a significant electric charge through a process similar to frictional charging.

Suppose a fly picks up a charge of +62 pC.
How many electrons does it lose to the surface it is walking across?
Physics
1 answer:
Umnica [9.8K]2 years ago
3 0

Answer:

Number of electrons, n=3.87\times 10^8\ electrons

Explanation:

Given that,

Charge on the fly, q=62\ pC=62\times 10^{-12}\ C

Let n is the number of electrons lose to the surface it is walking across. It is case of quantization of electric charge. It is given by :

q=ne

n=\dfrac{q}{e}

n=\dfrac{62\times 10^{-12}}{1.6\times 10^{-19}}

n = 387500000 electrons

or

n=3.87\times 10^8\ electrons

So, there are 3.87\times 10^8 electrons. Hence, this is the required solution.

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A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o
olya-2409 [2.1K]

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

  m=8/32=0.25

To find the spring constant

Kx=mg    (given that c=6 in and mg=8 lb)

K(0.5)=8               (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F  

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0
2 years ago
Can a body possess velocity at the same time in horizontal and vertical directions?​
iogann1982 [59]

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

5 0
1 year ago
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
patriot [66]

If I have done my math correctly, your answer is 67.5

4 0
2 years ago
Read 2 more answers
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

3 0
2 years ago
A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

4 0
2 years ago
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