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2 years ago

5

Two small plastics balls are hanging on silk threads, 50 mm from each other. The charges on the balls are indicated 1.3.1 Calcul

ate the magnitude of the force that the balls iniatially exert on each other

1 answer:

Nastasia [14]2 years ago

5 0

I calculated the magnitude of the force that the balls imitially exert to each other

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I need help please, thank you!!!

stiks02 [169]

Answer:

135 mph

because 7+9=16

8 is left

if 7=40

9=50

then 8=45

add 40+50+45mph

=135

4 0

2 years ago

A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material

denis-greek [22]

The radius of the cylinder is equal to half the diameter:

$r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm$

The volume of the cylinder is given by:

$V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3$

where h is the heigth of the cylinder. Converting into meters,

$V=1.28 \cdot 10^{-4} m^3$

And the density of the material will be given by the ratio between the mass and the volume:

$d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3$

5 0

3 years ago

In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a

ohaa [14]

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

= 0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

6 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

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

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago

You run<br> completely around a 400m track in<br> 80s. What was your average velocity?

dalvyx [7]

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

6 0

3 years ago