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2 years ago

What speed should a satellite of mass 4,900 kg moving around

1 answer:

Rudiy272 years ago

8 0

Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.

<u>Given the following data:</u>

Gravitational constant = 6.67 × 10⁻¹¹ m/kg²

Mass of Moon = 7.36 × 10²² kg

Distance, r = 4.2 × 10⁶ m.

<h3>How to determine the speed of this satellite?</h3>

In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.

This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:

Fc = Fg

mv²/r = GmM/r²

<u>Where:</u>

m is the mass of the satellite.

M is mass of the Moon.

Making v the subject of formula, we have;

v = √(GM/r)

Substituting the given parameters into the formula, we have;

v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)

v = √(1,168,838.095)

v = 1,081.13 m/s.

Speed, v = 1.8 × 10³ m/s.

Read more on speed here: brainly.com/question/20162935

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A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago

What is the direction of the magnetic field at point Z?

Tatiana [17]

The direction of the magnetic field at point Z ; INTO THE SCREEN

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3 years ago

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

$P.E=-\dfrac{GmM}{R+h}$

The kinetic energy at height above the surface of the earth

$K.E = 0$

The total energy at height above the surface of the earth

$E = K.E+P.E$

$E = -\dfrac{GmM}{R+h}$ ....(I)

The total energy at the surface of the earth

$E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$ ....(II)

We need to calculate the speed of the object just before it hits Earth

From equation (I) and (II)

$-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

Here, h = R

$-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

$v= \sqrt{\dfrac{GM}{R}}$

Hence, The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$ .

4 0

3 years ago

Gravity is a force of attraction that acts

bixtya [17]

Answer:gravity is a force of attraction that acts between two masses separated by distance

Explanation:

Gravity is a force of attraction that acts between two masses separated by distance

5 0

3 years ago

A Carnot engine, operating between hot and cold reservoirs whose temperatures are 800.0 and 400.0 K, respectively, performs a ce

nalin [4]

Answer:

The ratio between W2 and W1 is

p=1.5

Explanation:

The W2/W1 ratio equals the efficiency ratio, since the heat input is constant.

Initially, the Carnot efficiency is

1 - (Tc/Th)

1 - 400/800 = 0.50

After lowering Tc, the efficiency becomes

1 - (Tc/Th)

1 - 200/800 = 0.75

So to determine the ratio between W2 and W1

W2/W1 = 0.75/0.50 = 1.5

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3 years ago