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3 years ago

Which particle determines the type of element you have.

Physics

2 answers:

aivan3 [116]3 years ago

7 0

The number of protons in an atom determines the type of element you have.

Lilit [14]3 years ago

3 0

All atoms have a dense central core called the atomic nucleus. Forming the nucleus are two kinds of particles: protons, which have a positive electrical charge, and neutrons, which have no charge. All atoms have at least one proton in their core, and thenumber of protons determines which kind of element an atom is.

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A 250 kg engine block is being dragged across the pavement by a sturdy chain with a tension force of 10,000 N, as seen in the fr

Musya8 [376]

Answer: B

Explanation:

The friction force is less than the tension force

4 0

3 years ago

The number of protons is equal to the atomio mass minus the alomo<br> number<br> A True<br> B. False

Agata [3.3K]

The answer is false. the amount of protons is the same amount as the mass.

6 0

3 years ago

The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 Km/s.

tino4ka555 [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) the mass of the star Alpha is $M_{\alpha}$ = $7.80*10^{29} kg$

b) the mass of the star Beta is $M_{\beta}$ = $2.34*10^{30}kg$

c) the radius of the orbit of the orange star $R_{0}$ = $1.9*10^{9}m$

d) the radius of the orbit of the black hole $R_{B} = 34*10^{8}m$

e) the orbital speed of the orange star $V_{0} = 4.4*10^{2}km/s$

f) the orbital speed of the black hole $V_{B} =77 km/s$

Explanation:

The generally formula for orbital speed is given as $v =\frac{2\pi R}{T}$

where v is the orbital speed

R is the radius of the star

T is the orbital period

From the question we are given that

alpha star has an orbital speed of $V_{\alpha} = 36km/s = 36000m/s$

beta star has an orbital speed of $V_{\beta} = 12km/s = 12000m/s$

the orbital period is $T = 137d = 137(86400)sec$ 1 day is equal 86400 seconds

Making R the subject of formula we have for the radius of the alpha star as

$R_{\alpha} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(36000m/s)}{2\pi}$

$=6.78*10^{10}m$

for the radius of the Beta star as

$R_{\beta} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(12000m/s)}{2\pi}$

$= 2.26*10^{10}m$

Looking at the value obtained for $R_{\alpha} and R_{\beta}$

Generally the moment about the center of the mass are equal then

$M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$

Thus $3M_{\alpha} = M_{\beta} -------(1)$

Generally the formula for the orbital period is given as

$T =\frac{2\pi(R_{\alpha+R_{\beta}})^{3/2 }}{\sqrt{G(M_{\alpha}+M_{\beta})} }$

Then

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Where G is the gravitational constant given as $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(6.78*10^{10}m +2.26*10^{10}m)^{2}}{(137d*86400s/d)^{2}(6.67*10^{-11}m^{3}kg^{-1}s^{-2})}$

$M_{\alpha} + M_{\beta} = 3.12*10^{30}kg -------(2)$

Thus solving (1) and (2) equations

Mass of alpha star is $M_{\alpha}=7.80*10^{29}kg$

and the Mass of Beta is $M_{\beta} =2.34*10^{30}kg$

Considering the equation

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Making $R_{\alpha} + R_{\beta}$ the subject

$R_{\alpha}+ R_{\beta} =\sqrt[3]{\frac{(M_{\alpha+M_{\beta}})T^{2}G}{4\pi^{2}} } -------(3)$

and considering this equation $M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$ from above

we have that $R_{\beta} =\frac{M_{\alpha}R_{\alpha}}{M_{\beta}}$

Considering Question C

Let the Orange star be denoted by (0) and

Let the black-hole be denoted by (B)

And we are told from the question that

Mass of orange star $M_{0} = 0.67M_{sun}$ and

Mass of black hole $M_{B} =3.8M_{sun}$

And mass of sun is $M_{sun} = 1.99*10^{30}kg$

Then $R_{B} = [\frac{0.67M_{sun}}{3.8M_{sun}} ]R_{0} =0.176R_{0}--------(4)$

We are also given that the period is $T =7.75 days = 7.75 (86400s)$

Considering equation 3

$R_{0} + 0.176R_{0} = \sqrt[3]{\frac{(0.67+3.8)(1.99*10^{30}kg)(7.75(86400s))^{2} 6.67*10^{-11}Nm^{2}/s^{2}}{4\pi ^{2}} }$

Thus for V616 Monocerotis, $R_{0} =1.9*10^{9}m$

Considering equation 4

The black-hole is

$R_{B}= 0.176R_{0}= 0.176*1.9*10^9 =34*10^8m$

From the formula for velocity of $V_{0} = \frac{2\pi R_{0}}{T} = 4.4*10^{9}km/s$

the velocity of $V_{B} = \frac{2\pi R_{B}}{T} =77km/s$

8 0

3 years ago

Air enters the diffuser of a jet engine operating at steady state at 18 kPa, 216 K and a velocity of 265 m/s, all data correspon

earnstyle [38]

Answer:

45.44m/s

Explanation:

To solve this problem it is necessary to go back to the concepts related to the first law of thermodynamics,

in which it deepens on the conservation of the Energy.

The first law of Thermodynamics is given by the equation:

$0 = \dot{Q}-\dot{W}+\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})+\dot{m}g(z_1-z_2)$

Where,

$\dot{Q}=$ Heat transfer

$\dot{W} =$ Work

$\dot{m} =$ Flow mass

$V_i =$ Velocity

$h_i =$ Specific Enthalpy

$g =$ Gravity

$z_i =$ Height

From this equation we have that there is not Heat transfer, Work and changes in Height. Then,

Then our equation would be,

$0=\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})$

Solving for $V_2,$

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the tables of ideal gas (air) at 216K we have,

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

From the tables at 250K, we have that

$h_2 = 250.05kJ/kg$

The velocity was previously given, then

$V_1 =265m/s$

Replacing in the equation:

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}$

$V_2 = 45.44m/s$

Therefore the velocity of the air at the diffuser exit is 45.44m/s

3 0

4 years ago

Describe the transformation of mechanical energy in a ball when it is thrown straight up and then comes back down

zhenek [66]

I believe that the mechanical energy would transform from starting out as kinetic, the reaching the top it would be potential, then go back to kinetic as it is falling back down.

I'm not 100% sure that this is right but if I had to take a guess this is what I would say.

6 0

3 years ago