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scoundrel [369]
3 years ago
11

The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on

the planet’s surface? Round your answer to the nearest whole number. m/s
Physics
2 answers:
rewona [7]3 years ago
8 0

Answer:

241 m/s

Explanation:

Converting the last answer from kilometers/hour to meters/second

sammy [17]3 years ago
6 0

Answer: First, we determine the circumference of the Mars by the equation below.  

                                     C = 2πr

Substituting the known values,

                                     C = 2(π)(3,397 km) = 6794π km

To determine the tangential speed, we divide the circumference calculated above by the time it takes for Mars to complete one rotation and that is,

                   tangential speed = 6794π km / 24.6 hours = 867.64 km/h

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Answer:

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Explanation:

{ \sf{Pasacal \: ( Pa) =  \frac{newtons}{metres {}^{2} } }} \\  \\ { \sf{Pasacal  \: (Pa) =  \frac{kg \times  {ms}^{ - 2} }{ {m}^{2} } }}

4 0
3 years ago
5. In an AC circuit the applied emf e = E0​
Flura [38]
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3 0
3 years ago
Six artificial satellites circle a space station at a constant speed. The mass m of each satellite, distance L from the space st
damaskus [11]

Answer:

The answer to the question is;

Based on their acceleration the rank of the satellites from largest to smallest is.

B   >→   A   >→    E   >→    C    >→     F    >→     D.

Explanation:

Acceleration is given by \frac{v^2}{r}

Therefore the acceleration for each of the satellite is given by

Satellite A)      \frac{160^2}{5000} =    5.12 m/s²

Satellite B)      \frac{160^2}{2500}  =   10.24 m/s²

Satellite C)      \frac{80^2}{2500}   =  2.56 m/s²

Satellite D)      \frac{40^2}{10000}  =  0.16 m/s²

Satellite E)       \frac{120^2}{5000}   =  2.88 m/s²

Satellite F)       \frac{80^2}{10000}  = 0.64 m/s²

Therefore in order of decreasing acceleration, from largest to smallest we have

Satellite B) > Satellite A) >Satellite E) >Satellite C)>Satellite F)>Satellite D).

3 0
3 years ago
Read 2 more answers
A child standing on a bus remains still when the bus is at rest. when the bus moves forward and then slows down, the child conti
kati45 [8]
I'm pretty sure the answer to this question is "newton's first law of motion" or d. Hope this helps. :)

4 0
4 years ago
An l-c cirucit with a 70 mh inductor and a ,54 F capacitor oscillates. The maximum charge on the capacitor is 11.5 C. What are t
Oxana [17]

This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

5 0
3 years ago
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