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raketka [301]
3 years ago
10

In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of

mechanical energy does the fuel get transformed?
Physics
1 answer:
nexus9112 [7]3 years ago
6 0

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

F_g = F_c

\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}

now we can say that kinetic energy of satellite is  given as

KE = \frac{1}{2}M_s v^2

KE = \frac{GM_sM_p}{2r}

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

U = -\frac{GM_sM_p}{r}

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

You might be interested in
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
USPshnik [31]

Answer:

Temperature increase = 2.1 [C]

Explanation:

We need to identify the initial data of the problem.

v = velocity of the copper sphere = 40 [m/s]

Cp = heat capacity = 387 [J/kg*C]

The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:

E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT  \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]

8 0
3 years ago
A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle
Contact [7]
<h2>Answers: </h2>

1) 1.359, 1.403

2) 2.207(10)^{8}m/s,  2.138(10)^{8}m/s    

Explanation:

The described situation is known as Refraction.  

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index n is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum (c=3(10)^{8}m/s) and the speed of light v in the second medium:

n=\frac{c}{v}   (1)

On the other hand we have the Snell’s Law:  

n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})   (2)  

Where:  

n_{1} is the first medium refractive index . We are told is the air, hence n_{1}\approx 1

n_{2} is the second medium refractive index  

\theta_{1} is the angle of the incident ray  

\theta_{2} is the angle of the refracted ray  

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and \theta_{1}=57.0\º   \theta_{2}=38.1\º:

(1)sin(57.0\º)=n_{2}sin(38.1\º)  

Finding n_{2}:

n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}  

n_{2}=1.359   (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and \theta_{1}=57.0\º   \theta_{2}=36.7\º:

(1)sin(57.0\º)=n_{2}sin(36.7\º)  

Finding n_{2}:

n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}  

n_{2}=1.403   (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

n_{red}=\frac{c}{v_{red}}

v_{red}=\frac{c}{n_{red}}

Substituting (3) in this equation:

v_{red}=\frac{3(10)^{8}m/s}{1.359}

v_{red}=2.207(10)^{8}m/s >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

n_{violet}=\frac{c}{v_{violet}}

v_{violet}=\frac{c}{n_{violet}}

Substituting (4) in this equation:

v_{violet}=\frac{3(10)^{8}m/s}{1.403}

v_{red}=2.138(10)^{8}m/s >>>>Speed of violet light

3 0
3 years ago
How fast must a 1000 kg car be moving to have a kinetic energy of 2.0*10^3
lawyer [7]
Kinetic Energy is defined by Ke=1/2mv^2. Plug in and solve for v.

2,000 = 1/2(1000)(v)^2
4=(v)^2
v=2 m/s

The car must move at 2 m/s to have a Ke of 2,000 Joules.
6 0
3 years ago
The width of the central maximum is defined as the distance between the two minima closest to the center of the diffraction patt
Gnom [1K]

Answer:

The value of the angle is \bf{ \sin^{-1}[h/am_{e}v]}.

Explanation:

Given:

The condition for diffraction minima is

a \sin \theta = m \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

where, a is the slit-width, \theta is the angle of incidence, m is the order number and \lambda is the wavelength of the light.

The wavelength of an electron traveling through a medium is governed by de Broglie's hypothesis.

According to de Broglie's hypothesis

\lambda &=& \dfrac{h}{p}\\               &=& \dfrac{h}{m_{e}v}

Here, h is Planck's constant, m_{e} is the mass of the electron and v is the velocity of the electron.

For first minimum m = 1.

From equation (1), we have

&& a \sin \theta = \dfrac{h}{m_{e}v}\\&or,& \theta = \sin^{-1}[\dfrac{h}{am_{e}v}]

6 0
3 years ago
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