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3 years ago

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If the mass of a portion is 1.67 x 10-27kg and the mass of an electron is 9.11 x 10-31kg, calculate the force of gravitation bet

ween: i. a proton and an electronii. two electronsiii. two protons. Take G= 6.67 x 10-11 Nm2kg-2, distance between the protons = 4.0m., distance between the electrons = 2 x 10-2m, distance between the proton and the electron = 5.4 x 10-11m

1 answer:

emmainna [20.7K]3 years ago

7 0

Answer:

1.38 into 10 power minus 58 Newton is the answer

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Object 1 has a mass of 3m and is moving to the right at a velocity of

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Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

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

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3 years ago

Martino travels from her house to her moms office building from there she walks to school after school she walks to the library

Jlenok [28]

Answer:

A) 0 miles

Explanation:

The displacement is the distance between the starting point and the end point of the route.

In this case, even if Matino takes a whole tour around the city, since<u> it ends in the same place where it started</u>, the difference between the starting and finishing point is zero, so its total displacement is zero.

Care must be taken to distinguish the terms of displacement and distance traveled, beacause they are not the same, since in this case the distance traveled would be 3.15miles, but the displacement is zero, because it ends at the point where it started.

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3 years ago

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Answer:

The sphere returns back to the neutral.

Explanation:

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When the positively charged rod is removed from the faf end of rod A, the electrons return to their original state and the charges redistributes themselves and a neutral state is re-established .

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3 years ago

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Lina20 [59]

Answer:

The beauty of science is everywhere.

Explanation:

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make sure to ask if you need anymore examples or proof :)

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2 years ago