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3 years ago

Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.

1 answer:

muminat3 years ago

3 0

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

$V = \frac{kq}{r}$

k = Coulomb's constant

q = Charge

r = Distance between them

$q = 18 pC \rightarrow q = 1.8*10^-11 C$

$d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m$

Replacing,

$V = \frac{kq}{r}$

$V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}$

$V = 135 V$

Therefore the potential at the surface of the raindrop is 135 V

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An electric fan has the power output of 60W. How much work is done if the fan operates for 120s?

Digiron [165]

Since power = work done/time, 60= work done/120, work done = 120*60 = 7200. So,work done = 7200N (Newton).
I'm not sure if you're supposed to convert the seconds to time.

4 0

4 years ago

block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th

Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0

4 years ago

– A man walks 50 meters due west, then 50 meters 60 north of west. Determine the magnitude and direction of the resultant disp

Fynjy0 [20]

Answer:

86.6, 45°

Explanation:

The diagram explains better.

Using vector component method:

We find the x and y components of the vectors :

For the first:

A = -50cos(0)i + 50sin(0)j

A = -50i

For the second:

B = -50cos(60)i + 50sin(60)j

B = -25i + 43.3j

The resultant vector is :

R = A + B

R = -50i - 25i + 43.3j

R = -75i + 43.3j

The magnitude is:

R = [(-75)² + (43.3)²]^½

R = 86.6m

The angle is

tanθ = (50/50) = 1

θ = 45°

5 0

3 years ago

Read 2 more answers

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

4 years ago

A marble rolling at a speed of 2.71 m/s falls of the end of a table that is 1.25 m high. How far from the base of the table does

katrin [286]

Answer:

The marble lands at a distance of 1.36 m from the base of the table.

Explanation:

The motion of the marble falling off the table is a projectile with initial velocity in the horizontal direction only.

The motion can be solved in two directions, the horizontal and vertical direction.

Along the vertical direction, the initial velocity is 0 m/s as it has only horizontal component initially. Also acceleration in the vertical direction is acceleration due to gravity.

Let us use equation of motion in vertical direction.

$y-y_0=v_{0y}t+\frac{1}{2}at^2$

Where,

$v_{0y}\rightarrow \textrm{vertical component of the initial velocity}.\\y\rightarrow \textrm{final position of the marble}\\y_0\rightarrow \textrm{initial vertical position}\\a_y\rightarrow \textrm{acceleration in the vertical direction}\\t\rightarrow \textrm{time taken to reach bottom}$

Now, plug in 0 for $y$ , $1.25$ for $y_0$ , 0 for $v_{0y}$ , -9.8 for $a_y$ . This gives,

$0-1.25=0+\frac{1}{2}(-9.8)t^2\\-1.25=-4.9t^2\\t^2=\frac{-1.25}{-4.9}\\t^2=0.255\\t=\sqrt{0.255}=0.501\ s$

Therefore, time to reach bottom is 0.501 s.

Now, consider the horizontal motion. There is no acceleration in the horizontal direction. So, distance is given as the product of horizontal velocity and time taken.

Horizontal distance covered by the marble is given as:

$x=v_{0x}\times t=2.71\times 0.501=1.36 \textrm{ m}$

Therefore, the marble lands at a distance of 1.36 m from the base of the table.

6 0

3 years ago