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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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Please help asap!!!!!!!!!!!
Tomtit [17]
The answer would be 20000

The answer in standard form/scientific notation would be 2 x 10^4

(The exponent is 4 because that's how many digits after the 2 there is)
8 0
2 years ago
A barge floating in fresh water (rho = 1000 kg/m^3) is shaped like a hollow rectangular prism with base area A = 550 m^2 and hei
patriot [66]

Answer:

Explanation:

A )

When empty , H₀ length of barge is inside water .

volume of barge inside water = A x H₀

Weight of displaced water = AH₀ x ρ x g

Buoyant force = weight of displaced water = AH₀ ρg

B)

It should balance the weight of barge

Weight = buoyant force

Weight = AH₀ ρg

mass of barge = weight / g

weight / g = AH₀ ρ

= 550 x .55 x 1000

= 302500 kg

6 0
3 years ago
Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction
Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity  is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension  force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0  

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁     ,    T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

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What is kinematics...? anyone on?​
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Answer:

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