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maw [93]
3 years ago
15

Calculating Orbital Period The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed o

f 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m.
Physics
2 answers:
arsen [322]3 years ago
5 0
The answer is, 5630 seconds.
Vadim26 [7]3 years ago
5 0

Answer:

Orbital period, T = 93.89 minutes

Explanation:

It is given that,

The tangential speed of the Hubble Space Telescope, v = 7750 m/s

The radius of earth, r=6.38\times 10^6\ m

We need to find the orbital period of the Hubble Space Telescope that orbits 569 km above the Earth's surface i.e. h = 569 km = 569000 m

The orbital speed is given by,

v=\dfrac{2\pi R}{T}

T is the orbital period

R = r + h

R = 6949000 m

T=\dfrac{2\pi R}{v}

T=\dfrac{2\pi \times 6949000\ m}{7750\ m/s}

T = 5633.78 s

or

T = 93.89 minutes

So, the orbital period of the Hubble Space Telescope is 93.89 minutes. Hence, this is the required solution.

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we see a magnet exerting a force on a current-carrying wire. Does a current-carrying wire exert a force on a magnet? Why or why
max2010maxim [7]

Answer:

Yes

Explanation: Electric and magnetic field are known to be inter-related, this implies that for any current carrying conductor there is a resulting magnetic field around the wire ( for example a current carrying conductor deflects a compass) and a magnetic field has been known to produce some amount current based on the<em> </em>principle of electromagnetic induction by Micheal Faraday.

The strength of magnetic field generated by a current carrying conductor is given by Bio-Savart law (purely mathematical) which is

B =\frac{u_{0}I }{2πr}

B= strength of magnetic field

I =current on conductor

r = distance on any point of the conductor relative to it center

If a current carrying could generate this magnitude of magnetic field, thus this magnetic field has the ability to interact (exert a force on any magnetic material) with any other magnetic material including a magnet.

Yes, a current carrying conductor can exert a force on a magnetic field

7 0
3 years ago
Read 2 more answers
Refrigerant-134a enters the expansion valve of a refrigeration system at 160 psia as a saturated liquid and leaves at 30 psia. D
KatRina [158]

Answer:

Temperature : 92.9 F

Internal Energy change: -2.53 Btu/lbm

Explanation:

As

mh1=mh2

h1=h2

In table A-11 through 13E

p2=120Psi, h1= 41.79 Btu/lbm,

u1=41.49

So T1=90.49 F

P2=20Psi

h2=h1= 41.79 Btu/lbm

T2= -2.43F

u2= 38.96 Btu/lbm

T2-T1 = 92.9 F

u2-u1 = -2.53 Btu/lbm

3 0
3 years ago
The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu
BartSMP [9]

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

8 0
3 years ago
URGENT
Advocard [28]

Answer:

give \\ mass(m) = 1350kg  \\ acceleration(a) = 1ms {}^{ - 2}  \ \\ sln\ \\  from \: our \: formula \\  \: f = ma \\1350kg \times 1ms {}^{ - 2}  \\ f = 1350newton

the force you applied to your car =1350N

5 0
3 years ago
Accelerations are produced by equal forces or unequal forces?
sukhopar [10]

Answer: unequal forces

Explanation: in order for something to accelerate it must speed up or slow down .  When unequal forces react it casuse a change in motion which is also know as acceleration .

5 0
3 years ago
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