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Ganezh [65]
2 years ago
9

Two small, positively charged spheres have a combined charge of 5.0 x 10 -5 C. If each sphere is repelled from the other by an e

lectrostatic force of 1.0 N when the spheres are 2.0 m apart, what is the charge, in micro-coulomb, on the sphere with the smaller charge?
Physics
1 answer:
geniusboy [140]2 years ago
4 0

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

q_1+q_2=5\times 10^{-5}\ C.............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

F=k\dfrac{q_1q_2}{d^2}

q_1q_2=\dfrac{F.d^2}{k}

q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}

q_1q_2=4.45\times 10^{-10}\ C............(2)

On solving the system of equation (1) and (2) using graph we get,

q_1=0.0000384\ C=38.4\ \mu C

q_2=0.0000116\ C=11.6\ \mu C

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

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zalisa [80]
You need 5 blocks of the smaller object to contain the same amount of volume of the bigger object

3 0
3 years ago
Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the
marysya [2.9K]

Volume of balloon = \frac{4}{3} \pi r^{3}

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V =  68+ 0.178*V

   0.934 V = 68

     V = 72.81 m^3

    \frac{4}{3} \pi r^{3}[/tex]= 72.81

     r = 2.59 m

So radius of helium balloon = 2.59 m

4 0
3 years ago
About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, a
Maurinko [17]

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

v^2=u^2+2gs

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

0=u^2-2\times9.8\times11.0

u=\sqrt{2\times9.8\times11.0}

u=14.68\ m/s

Hence, The speed of the water is 14.68 m/s.

7 0
3 years ago
Find the unit vector in the direction of vector B=4i+2j-5k​
solmaris [256]

Answer:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

Explanation:

for the unit vector, we need to divide the given vector by its norm, because it should be in the SAME direction as the original vector, but of magnitude "1".

We notice that the norm of the given vector is:

\sqrt{4^2+2^2+(-5)^2} =\sqrt{45}

Then, the unit vector becomes:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

7 0
2 years ago
a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra
Nikolay [14]

The dimensions of the rectangle are:

l = 50 m

b = 100/\pi m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is =\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l

Therefore,  \pi b+2l=200

b=(200-2l)/\pi

Now,

Area, A = l(200-2l)/\pi=(200l-2l^{2} )/\pi

Now, differentiate A w.r.t l:

Again differentiating w.r.t 'l', we get:

d^{2} A/dl^{2} =-4l/\pi< 0

Thus we get maximum are when dA/dl=0

Therefore,

(200-4l)/\pi=0

l = 50 m

Now, from

\pi b+2l=200

\pi b=200-2*50

b=100/\pi

r=b/2=50/\pi

To know more about area of a recatangle, visit the link

brainly.com/question/20693059

#SPJ4

4 0
1 year ago
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