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3 years ago

How are scientists able to measure the age of earth?

Physics

1 answer:

RoseWind [281]3 years ago

4 0

Explanation:

Through the uranium-lead dating, you can establish the age of rocks that formed from 1 million years ago to 4.5 billion years with enough precision. Dating meteorites using this method, It is possible to establish the age of the earth, since these are pieces that are present since the formation of the solar system, so they are approximately the same age as the Earth.

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

Read 2 more answers

A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ

Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0

2 years ago

A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?

lubasha [3.4K]

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n ≈ 14.0 g/mol

4 0

2 years ago

A line from the Brackett series of hydrogen has a wavelength of 1945nm (or 1.0979x10^7m). From which state did the electron orig

Zarrin [17]

We use the Rydberg Equation for this which is expressed as:

1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]

where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m. Thus,

1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]

Solving for n2, we obtain n=1.

5 0

3 years ago

An airplane has an effective wing surface area of 19.4 m2 that is generating the lift force. In level flight the air speed over

alexgriva [62]

Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

density of air = 1.3 kg/m³

weight of plane

From Bernoulli's principle

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2$

where 1 and 2 are two different locations at the same geo potential level

so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)$

$\Delta P =1227.2\ N/m^2$

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

7 0

3 years ago