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2 years ago

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The Earth's escape speed (the speed you need to get away forever) is about 40,000 kilometers per hour. Escape speed depends on t

he gravity of the object trying to hold the spacecraft from escaping. Based on your understanding of gravity, how will the escape speed from the Moon compare to the escape speed from Earth

1 answer:

Anvisha [2.4K]2 years ago

5 0

The Moon s escape speed will be smaller than Earth's.

What is escape speed:

The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.

The formula of escape speed is

$v = \sqrt{\frac{2GM}{R} }$

where

v is escape velocity

G is universal gravitational constant

M is mass of the body to be escaped from

r is distance from the center of the mass

we can say that,

Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.

we know that,

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.

since, v ∝ g

The Moon s escape speed will be smaller than Earth's.

Learn more about escape speed here:

<u>brainly.com/question/15318861</u>

#SPJ4

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What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0

2 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

2 years ago

Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li

tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

W=9.81x Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: $(F_{J} +F_{T} =9.81x n)$

$F_{T} (510)=9.81x$ (210+21.5)

$F_{T} (510)=9.81 x (231.5)$

$F_{T} =4.452x N.$

$F_{J} =9.81x-4.452x$

$F_{J} =5.358 xN$

Therefore $F_{J} \geq F_{T}$

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

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1 year ago

TRUE/FALSE: Water balance is the balance between water intake and water excretion?

Alex787 [66]

True, because water balance is the balance between intake and output

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3 years ago