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elixir [45]
3 years ago
15

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

us 8.20×10^7 m, and orbital speed 4800 m/s. The second satellite has mass 54.0 kg and orbital radius 7.00×10^7 m.
A) What is the orbital speed of this second satellite?
v=?
Physics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}

Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

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Answer:

Explained

Explanation:

Displacement is the change in the position of an object with reference to a starting point. for displacement to occur the position of the object must change.

Here the bacteria although moving but it is moving back and forth, meaning its initial and final positions are the same and hence no displacement. Whereas distance is the total distance traveled no matter in what direction. Hence, The total distance traveled by a bacterium is large for its size, while its displacement is small.

4 0
4 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10 nm. Assume that the walls act li
ser-zykov [4K]

Answer:

E = 1.1 10⁶ N / C

Explanation:

In this case they indicate that we can approximate the membrane as a parallel plate capacitor, we can use

            E = \frac{\sigma}{\epsilon_o }

note that in this case the electric field created by each plate goes in the same direction, they are added

let's calculate

            E =  \frac{10^{-5}}{8.85 \ 10^{-12}}

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3 0
3 years ago
A wave with an amplitude of 9.3 mm is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. I
dsp73

Answer:

The rate of transfer of energy is equal to 23.76W or 23.76J/s as may be required both forms are correct. The physical quantities needed to calculate the rate of energy transfer are the linear mass density or mass per unit length, tension force, amplitude, angular frequency( which is equal to 2pi •f )

Explanation:

The required quantity is the average power or average rate of energy transfer which differs from the maximum or instantaneous rate of energy transfer. The calculation steps to the answer above can be found in the attachment below. Should the requested quantity be the instantaneous quantity the answer will be 2 x Pav which equals 47.52W or 47.52J/s.

3 0
3 years ago
The table shows the specific heat capacities of various substances. How much energy is required to raise the temperature of 5g o
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Answer:

Q = 50.25 [J]

Explanation:

To solve this problem we must use the following equation that relates the temperature change with the mass and with the specific heat.

Q = m*Cp*(DT)

where:

Q = energy in form of heat [J]

m = mass = 5 [g] = 0.005 [kg]

Cp = specific heat = 1005 [J/kg*°C]

DT = temperature change = 10 [°C]

Now replacing:

Q = 0.005*1005*10

Q = 50.25 [J]

4 0
4 years ago
Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of
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To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

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Our values are given as

P = 1atm

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Using the equation to find n, we have:

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n = \frac{PV}{RT}

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n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

7 0
3 years ago
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