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Mamont248 [21]
3 years ago
9

An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg

e Q is brought close to, but not touching, sphere A. Describe what happens to the charges on A and B as they are brought close but not touching. If we now remove sphere B and place it far away, what is the charge on sphere A
Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

0

Explanation:

  • If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
  • As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:

       Q +(-Q) =0.

  • If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
  • The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.
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Calculate the minimum frequency of ultrasound (in Hz) that will allow you to see details as small as 0.193 mm in human tissue. (
STatiana [176]

Answer:

f = 7.97 x 10⁶ Hz = 7.97 MHz

Explanation:

The speed of a wave is given by the following formula:

v = f\lambda

where,

v = speed of the ultrasound wave through human tissue = 1540 m/s

f = frequency of ultrasound wave required = ?

λ = wavelength of ultrasound waves = smallest detail required = 0.193 mm

λ = 0.193 mm = 1.93 x 10⁻⁴ m

Therefore,

1540\ m/s = f(1.93\ x\ 10^{-4}\ m)\\f = \frac{1540\ m/s}{1.93\ x\ 10^{-4}\ m}

<u>f = 7.97 x 10⁶ Hz = 7.97 MHz</u>

8 0
3 years ago
Calculate the area of a square with a length of 5cm
mihalych1998 [28]

Answer:

25cm^2

Explanation:

area of square = side × side

length of side given = 5

area of this square = 5× 5

= 25cm^2

hope it helps

6 0
3 years ago
Read 2 more answers
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
3 years ago
The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is
mestny [16]

Answer:

<em>3924 Pa</em>

<em></em>

Explanation:

Volume of cylinder = 2 L = 0.002 m^3  (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m  (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = \pi r^{2}

where \pi = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x 0.04^{2} = 0.005 m^2

<em>height h of the water in this cylinder = volume/area</em>

h = 0.002/0.005 = 0.4 m

<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>

where

p = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

h = height of water within this cylinder = 0.4 m

pressure = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>

3 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
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