Suppose an objects intai velocity is 10 mis and its firal velocity is 4 mis. Mas is constant

Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to for

hodyreva [135]

Answer:

$V = 228\ V$

Explanation:

given,

charge of two spherical drop = 0.1 nC

potential at the surface = 300 V

two drops merge to form a single drop

potential at the surface of new drop = ?

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

$r = \dfrac{9\times 10^9\times 0.1 \times 10^{-9}}{300}$

r = 0.003 m

volume = $2 \times \dfac{4}{3}\pi r^3$

= $2 \times \dfac{4}{3}\pi \times 0.003^3$

= 2.612 × 10⁻⁷ m³

$\dfac{4}{3}\pi R^3 = 2.612\times 10^{-7}$

$R =\sqrt[3]{\dfrac{2.612 \times 10^{-7}\times 3}{4\times \pi}}$

R = 0.00396 m

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

$V = \dfrac{9\times 10^9 \times 0.1 \times 10^{-9}}{0.00396}$

$V = 227.27$

$V = 228\ V$

6 0

3 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

3 years ago

Hi gir_ls join nkd-mbja-nuj

GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

3 0

2 years ago

GOOD POINTS HELP PLEASE IN SCIENCE

jeka94

The answer is c because a metallic bond Is 1. formed of the attraction between positively charged metal nuclei
2. and surrounding sea electrons

6 0

3 years ago

g An electric hot plate raises its own internal energy and the internal energy of a cup of water by 9000J, and there is at the s

DochEvi [55]

Answer:

11700j

Explanation:

add the two because the plate has to maintain the temp.

2700+9000=11700

3 0

2 years ago