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3 years ago

Help please what is the answer to this

Physics

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

1870 dalton simple sphere

1897 plum pudding positive mass........

1905 Rutherford electrons orbitting....

1913 Bohr electrons in shells......

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A body on the surface of a planet with the same radius as Earths weighs 10 times more than it does on Earth. What is the mass of

nadya68 [22]

Answer:

Explanation:

Given

radius of Planet is equal to radius of Earth

$r_p=r_e$

Weight of body on Planet $F_p=mg_p$

where m=mass of body

$g_p=acceleration\ due\ to\ gravity\ on\ surface\ of\ Planet$

Weight of body on earth $F_e=mg_e$

$g_e=acceleration\ due\ to\ gravity\ on\ Earth$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

where G=gravitational constant

M=mass of Planet

r=radius of planet

for earth $g_e=\frac{GM_e}{r_e^2}$

for planet $g_p=\frac{GM_p}{r_p^2}$

substituting these values in $F_e$ and $F_p$

$F_p=m\times \frac{GM_p}{r_p^2}---1$

$F_e=m\times \frac{GM_e}{r_e^2}---2$

divide 1 and 2

$\frac{F_p}{F_e}=\frac{m\times \frac{GM_p}{r_p^2}}{m\times \frac{GM_e}{r_e^2}}$

$10=\frac{M_p}{M_e}$

$M_p=10M_e$

6 0

3 years ago

In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf, having about the same mass as it doe

masya89 [10]

The formula of density is given by

Density = Mass ÷ Volume

We have:
Mass = 1.989 × 10³⁰ kg
Volume = $\frac{4}{3} \pi r^3$ = $\frac{4}{3} \pi (7500)^3 = 1.767*10^12$ km³

Density = $\frac{(1.989)(10^{30})}{(1.767)(10^{12})}$ =1.13×10¹⁸ kg/km³

Converting 1.13 × 10¹⁸ kg/km³ to g/cm³

1.13 × 10¹⁸ kg = 1.13 × 10¹⁸ × 10³ = 1.13 × 10²¹ grams
1 km³ = 1 × 10⁶ cm³

(1.13 × 10²¹) ÷ 10⁶ = 1.13 × 10¹⁵ gr/cm³

Answer: Density 1.13 × 10¹⁵ gr/cm³

8 0

4 years ago

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Andrew [12]

What are the following answers?

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3 years ago

A car of mass 1230 kg is on an icy driveway inclined at an angle of 39°. The acceleration of gravity is 9.8 m/s². θ If the incli

tamaranim1 [39]

Answer:

Acceleration of the car, $a=6.16\ m/s^2$