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2 years ago

A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?

Physics

1 answer:

11111nata11111 [884]2 years ago

7 0

Answer:

$\sigma=0.014\ C/m^2$

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

$\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2$

So, the surface charge density on the sphere is $0.014\ C/m^2$ .

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A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin

krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

$F=(0.850) sin (\frac{x}{2.00})$ [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

$a=\frac{F}{m}$

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

$a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00})$ [m/s^2]

When x = 3.14 m, the acceleration is:

$a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2$

Now we can find the final speed of the particle by using the suvat equation:

$v^2-u^2=2ax$

where

u = 8.00 m/s is the initial velocity

v is the final velocity

$a=5.67 m/s^2$

x = 3.14 m is the displacement

Solving for v,

$v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s$

And the speed is just the magnitude of the velocity, so 9.98 m/s.

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2 years ago

Consider a very small hole in the bottom of a tank 17.0 cm in diameter filled with water to a heightof 90.0 cm. Find the speed a

umka21 [38]

Answer:

Speed of water, v = 4.2 m/s

Explanation:

Given that,

Diameter of the tank, d = 17 cm

It is placed at a height of 90 cm, h = 0.9 m

We need to find the speed at which the water exits the tank through the hole. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 0.9}$

v = 4.2 m/s

So, the speed of water at which the water exits the tank through the hole is 4.2 m/s. Hence, this is the required solution.

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2 years ago

A student states that 203.9 g of table sugar will form a saturated solution in 100 g of water.

riadik2000 [5.3K]

The temperature and the solubility of sugar at that temperature

Explanation:

The amount of substance which can be dissolved in the solvent depends on the temperature.

As the temperature increases, more substance can be dissolved.

A solution is saturated if any more of the solute cannot be dissolved in the solution at the given temperature

Hence we need to know the temperature and also the amount of substance which can be dissolved(solubility) at the same temperature

a) the statement given in option A is correct

b) molar mass has no correlation with the substance's solubility and hence option b is not correct

c) The percent by volume of the solution is not needed to find if the solution is saturated and hence option c is not correct

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x=((12.3/100)m)cos[(1.26s^−1)t]
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C) electrical energy is transformed into heat energy

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