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4 years ago

8

If the radius of the atom is the distance from point A to D, where is the MOST likely location of the LEAST concentration of mas

s?

1 answer:

lana [24]4 years ago

3 0

Answer: B

Explanation:

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On a distant planet, the GPE of a 65 kg astronaut is 11,115 j when they are on a 46 m tall cliff. What is the acceleration due t

Ainat [17]

Answer:Learn what gravitational potential energy means and how to calculate it. ... a pulley and rope, so the force due to lifting the box and the force due to gravity, ... would be used by an elevator lifting a 75 kg person through a height of 50 m if the ... When you are close to a planet you are effectively bound to the planet by gravity ..

Explanation:

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3 years ago

PLEASE HELP 15 POINTS AND BRAINIEST!!!<br> Reporting fake answers

Zolol [24]

Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

$T = 2\pi \sqrt{\frac{r^3}{GM} }$

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

$F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N$

5 0

3 years ago

Future space stations could create an artificial gravity by rotating. Consider a cylindrical space station that rotates with a p

aleksandrvk [35]

Answer:

P = 2 pi R / v period of space station

F / m = v^2 / R centripetal force per unit of mass

So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2

Also, F / m = 9.8 m/s^2 earth's gravitational attraction

So 9.8 = 4 pi^2 R / P^2 or R = 9.8 P^2 / 4 * pi^2) = 195 m

Or D = 2 R = 390 m the diameter required

8 0

3 years ago

A force of 1 newton will cause a mass of 1 kg to have an acceleration of 1 m/s^2. If a force of 5 newtons is applied to a mass o

nignag [31]

For any force and any mass, the acceleration is always

(the strength of the force) divided by (the mass) .

If a force of 5 newtons is applied to a mass of 5 kg, the acceleration will be

(5 Newtons) / (5 kg) = <em>1 m/s²</em>

7 0

3 years ago

g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 Ã— 1026 W and assume that th

vagabundo [1.1K]

Answer:

The value is $N = 1.107 *10^{45 } \ photons$

Explanation:

From the question we are told that

The power output from the sun is $P_o = 4 * 10^{26} \ W$

The average wavelength of each photon is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

Generally the energy of each photon emitted is mathematically represented as

$E_c = \frac{h * c }{ \lambda }$

Here h is the Plank's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }$

=> $E_c = 3.614 *10^{-19} \ J$

Generally the number of photons emitted by the Sun in a second is mathematically represented as

$N = \frac{P }{E_c}$

=> $N = \frac{4 * 10^{26} }{3.614 *10^{-19}}$

=> $N = 1.107 *10^{45 } \ photons$

5 0

3 years ago