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4 years ago

Mercury is in the 80th position in the periodic table. How many protons does it have?

Physics

2 answers:

BaLLatris [955]4 years ago

8 0

mercury is 80 protons

aalyn [17]4 years ago

4 0

Mercury has 80 protons.

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Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0

3 years ago

Read 2 more answers

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

tatiyna

(a) $2.7\cdot 10^{25} kg$

The acceleration due to gravity on the surface of the planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we know:

$g=22.4 m/s^2$

$d=1.8\cdot 10^7 m$ is the diameter, so the radius is

$R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m$

So we can re-arrange eq.(1) to find M, the mass of the planet:

$M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg$

(b) $4.8\cdot 10^{31}kg$

The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$ (1)

where

m is the mass of the planet

M is the mass of the star

v is the orbital speed of the planet

r is the radius of the orbit

The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:

$v=\frac{2\pi r}{T}$

where

$T=402 days = 3.47\cdot 10^7 s$

Substituting into (1) and re-arranging the equation

$m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}$

And substituting the numbers, we find the mass of the star:

$M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg$

4 0

3 years ago

An instant soup is very low in fat and calories but high in sodium. Can this food be labeled “ healthy”?

Y_Kistochka [10]

Answer:

Explanation:

it is not healthy because sodium is not good to our body

5 0

3 years ago

An air tank of volume 1.5 m3 is initially at 800 kPa and 208C. At t 5 0, it begins exhausting through a converging nozzle to sea

Serggg [28]

Answer:

(a) m = 0.141 kg/s

(b) t = 47.343 s

Explanation:

Given that:

The volume of air in the tank V = 1.5 m³

The initial temperature in the tank is supposed to be 20° C and not 208 C;

So $T_o = 20^0 C = ( 20 +273) K = 293K$

The initial pressure in the tank $P_o= 800 \ kPa$

The throat area $A_t$ = 0.75 cm²

To find the initial mass flow in kg/s.

Lets first recall that:

Provided that $\dfrac{P_{amb}}{P_{tank}}< 0.528$ , then the flow is choked.

Then;

$\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}$

$\dfrac{P_{amb}}{P_{tank}}= 0.1266$

From what we see above, it is obvious that the ratio is lesser than 0.528, therefore, the flow is choked.

Now, for a choked nozzle, the initial mass flow rate is determined by using the formula:

$m = \rho \times A \times V$

where;

$\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}$

$\rho =9.51350( 0.8333 ) ^{2.5}$

$\rho =6.03 \ kg/m^3$

$T = T_o \bigg ( \dfrac{2}{k+1}\bigg)$

where;

$T_o = 293 \ K$

$T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)$

$T = 293 \bigg ( \dfrac{2}{2.4}\bigg)$

$T = 293 ( 0.8333)$

T = 244.16 K

During a critical condition when Mach No. is equal to one;

$V = a = \sqrt{kRT}$

$V = \sqrt{1.4 \times 287 \times 244.16}$

$V = \sqrt{98103.488}$

V = 313.214 m/s

Thus, the initial mass flow rate $m = \rho \times A \times V$

m = 6.03 × 0.75 × 10⁻⁴ × 313.214

m = 0.141 kg/s

(b)

The mass balance formula for the control volume surrounding the tank can be expressed as:

$\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)$

$= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m$

When the air mass flow rate is:

$m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

Thus; replacing $m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$ in the previous equation; we have:

$\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

$\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt$

Taking the differential of both sides from 0 → t

$In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t$

$In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t$

$\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$

So, when the pressure P = 500 kPa, the time required is:

$\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

t = 47.343 s

(c)

Let us recall that:

The choking on the nozzle occurred when $\dfrac{P_{amb}}{P_{tank}} = 0.528$

$\dfrac{P_{amb}}{0.528} = P_{tank}$

$\dfrac{101.35}{0.528} = P_{tank}$

$P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa$

From $\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$ ; the time required for $P_{tank} \simeq 192 \ kPa$ is:

$\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

By solving:

t = 143.745 s

7 0

3 years ago

What is the SI unit for acceleration?

qwelly [4]

The base unit for acceleration is m s ^-2

6 0

3 years ago