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nydimaria [60]
3 years ago
13

A satellite is one Earth radius above the surface of Earth. How does teh acceleration due to gravity at that location compare to

acceleration at the surface of Earth?
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

Explanation:

Given

Satellite is at distance equal to earth radius from earth surface

r_{eff}=r+r=2 r

Where r=radius of earth

acceleration due to gravity is given by

g=\frac{GM}{r^2}

for r=2 r

g'=\frac{GM}{(2r)^2}

g'=\frac{GM}{r^2}\times 0.25

g'=0.25 g

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A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The r
ehidna [41]

Answer:

Mass of the planet = 6.0 × 10^{24}

Explanation:

Time period = 2π (R + h) / v

Orbital speed (v) = √GM / (R + h)

T² = 4π² (R + h)² / (GM/ (R + h))

    = 4π² (R + h)³ / GM

  making m the subject of the formula

m = 4π² (R + h)³ / GT²

   = 4π² ( 6.38 × 10^{6} + 230 × 10³ )³ / ( 6.67 × 10^{-11}) × (89 × 60)²

    = 4π² ( 6610000)³ / ( 6.67 × 10^{-11}) × (89 × 60)²

    = 5.99 × 10^{24}

     = 6.0 × 10^{24}

5 0
3 years ago
g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Fin
schepotkina [342]

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

4 0
3 years ago
How much work is done when a 25-N horizontal force is applied to a 10-kg box, causing it to move 5 meters?
Katarina [22]

Answer:

W=50*sin(25)

Explanation:

W=F*D*sin(25)

W=10*5*sin(25)

W=50*sin(25)

8 0
3 years ago
A single slit is illuminated by light of wavelengths λa and λb, chosen so that the first diffraction minimum of the λa component
agasfer [191]

Answer:

λ_A = 700 nm ,   m_B = m_a 2

Explanation:

The expression that describes the diffraction phenomenon is

         a sin θ = m λ

where a is the width of the slit, lam the wavelength and m an integer that writes the order of diffraction

a) They tell us that now lal_ A m = 1

         a sin θ = λ_A

coincidentally_be m = 2

          a sin  θ = m λ_b

as the two match we can match

         λ _A = 2 λ _B

         λ_A = 2 350 nm

         λ_A = 700 nm

b)

For lam_B

       a sin  λ_A  = m_B  λ_B

For lam_A

        a sin θ_A = m_ λ_ A

to match they must have the same angle, so we can equal

           m_B  λ_B = m_A  λ_A

           m_B = m_A  λ_A / λ_B

           m_b = m_a 700/350

           m_B = m_a 2

8 0
3 years ago
Help on matching please, struggling on it. Even just 1 or 2 would help
n200080 [17]

Answer:

7. free fall -- h. 9.8m/s^2

3. Velocity -- x. 60 km/hr west

6. Acceleration -- d. change in velocity/time

8. Centrifugal --  s. towards the centre

13. Work done --w. Force * displacement

5. Uniform circular motion --j. spin cycle in washer

18. Power -- r. kW an hour

7. g -- a. 10N

hope this helps

6 0
3 years ago
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