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3 years ago

Can you help me please

Physics

1 answer:

Mrac [35]3 years ago

8 0

A denser liquid will sink to the bottom while a less denser liquid will float. Also, liquids composed of polar molecules such a water will not form chemical <span>bonds with nonpolar molecules.

Density.

hope this helps!
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A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t

Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

Electric field $E=7.80\times10^{2}\ N/C$

We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

We need to calculate the acceleration

Using newton's law

$F = ma$

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

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m

3 0

3 years ago

the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?

hichkok12 [17]

The pressure at the depth h in the ocean is given by (Stevin's law)
$p= p_0 + \rho g h$
where
$p_0 = 1.0 \cdot 10^5 Pa$ is the atmospheric pressure
and $\rho g h$ is the pressure exerted by the column of water of height h=4267 m, with $\rho = 1000 kg/m^3$ being the water density and $g=9.81 m/s^2$ .
Substituting, we find
$p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa$
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so $1 atm=1.0\cdot 10^5 Pa$ , therefore we just need to divide by this number:
$p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm$

7 0

3 years ago

A nucleus with a mass number of 64 has a mean radius of about: _________.

dangina [55]

Answer:

The correct option is A

Explanation:

From the question we are told that

The mass number is $A= 64$