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2 years ago

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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?

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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On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car c

Nana76 [90]

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

$F=\mu mg = m \frac{v^2}{r}$ (1)

where

$\mu$ is the coefficient of friction

m is the mass of the car

g = 9.8 m/s^2 is the acceleration due to gravity

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

$\mu=0.50\\v = 20 mph = 8.9 m/s$

So the radius of the curve is

$r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m$

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when $\mu=1.0$ . We find:

$v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph$

7 0

1 year ago

Read 2 more answers

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

1 year ago

A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centripetal force to remain

Artyom0805 [142]

Answer:

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

Explanation:

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2 years ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

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2 years ago

Rod A has twice the diameter of rod B, but both are made of iron and have the same initial length. Both rods are now subjected t

Vilka [71]

Answer:

ΔLa/ΔLb = 1

Explanation:

The change in length of a solid is given by the following formula:

ΔL = α L ΔT

where,

ΔL = Change in length

α = coefficient of linear expansion

L = Original Length

ΔT = Change in Temperature

Since, the length and change in temperature for both rods are same. Also, the material of each rod is same, which implies that coefficient of linear expansion for both rods is same. Hence, the ratio of change in length of both rods will be:

<u>ΔLa/ΔLb = 1</u>

4 0

2 years ago