¿De qué altura cae un cuerpo que tarda 4 s en llegar al suelo? usar g = 10 m/s2 *

Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon

Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

$T^2=\frac{4\pi}{GM}r^3$

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

$T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min$

the comet takes around 16.63min

8 0

3 years ago

A. The flow of energy from Earth's interior to the surface is about 50 terawatt (1 terawatt = 1e12 joule/sec). This is the energ

alisha [4.7K]

A) The amount of geothermal energy that reaches Earth's surface in one day is : 43.2 * 10¹⁷ joules

B) Humanity's current consumption of energy is; 71.4 GJ per individual.

C) Ratiois ; 60.5 * 10⁶ Joules

<u>Given data :</u>

Amount of energy flow = 50 terawatt = 50*10¹² joule/sec.

<h3>Determine the geothermal energy that reaches the earth</h3>

Since the geothermal energy reaches the earth in a single day

Energy = power * time

Time = 24 hours ( 1 day ) = 86400 secs.

Energy = 50*10¹² joule/sec * 86400 secs

= 43.2 * 10¹⁷ joules

<h3 /><h3>B) Estimating humanity's current consumption of energy in one day.</h3>

Humanity's current consumption of energy in one day is estimated to be 71.4 GJ per individual.

<h3>C) Determine the ratio of geothermal energy to human's current annual consumption </h3>

Geothermal energy / human current annual consumption

= 43.2 * 10¹⁷ Joules / 71.4 * 10⁹ Jolues

= 60.5 * 10⁶ Joules

Hence we can conclude that A) The amount of geothermal energy that reaches Earth's surface in one day is : 43.2 * 10¹⁷ joules

B) Humanity's current consumption of energy is; 71.4 GJ per individual.

C) Ratiois ; 60.5 * 10⁶ Joules

Learn more about Flow of energy : brainly.com/question/14427111

3 0

2 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago

Which activities demonstrate speed the most? <br> Look at pic

steposvetlana [31]

B: Sprinting. It is the best answer because it demonstrates speed better than the others

5 0

3 years ago

When energy from the Sun reaches the air above land, the air warms and rises. Along a coastline, cooler air above the ocean flow

Naily [24]

Answer:

convection

Explanation:

Convective heat transfer, often referred to simply as convection, is the transfer of heat from one place to another by the movement of fluids. this time fluid is out atmosphere.

3 0

3 years ago