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What is the station's orbital speed? the radius of earth is 6.37×106m, its mass is 5.98×1024kg.

Paha777 [63]

Answer:

$v_o=2503.08\frac{m}{s}$

Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

$v_o=\sqrt{\frac{GM}{r}}$

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

$v_o=\sqrt{\frac{6.674*10^{-11}\frac{m^3}{kg\cdot s^2}(5.98*10^{24}kg)}{6.37*10^6m}}\\v_o=2503.08\frac{m}{s}$

3 0

2 years ago

Read 2 more answers

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a

Nina [5.8K]

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'\dfrac{v}{v+v_s}\\\Rightarrow 1.01f'=f'\dfrac{v}{v+v_s}\\\Rightarrow 1.01=\dfrac{343}{343+v_s}\\\Rightarrow v_s=\dfrac{343}{1.01}-343\\\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'\dfrac{v}{v+v_s}\\\Rightarrow 0.99f'=f'\dfrac{v}{v+v_s}\\\Rightarrow 0.99=\dfrac{343}{343+v_s}\\\Rightarrow v_s=\dfrac{343}{0.99}-343\\\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

3 0

2 years ago

Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 N

solong [7]

Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10 (g = 10 m/ $s^{2}$ )

m = $\frac{25}{10}$

= 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.

5 0

2 years ago

What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each

suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
A = area of plates and
d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0

2 years ago

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

alexdok [17]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

$X=V_{o}*t + \frac{a*t^{2}}{2}$ Replacing the known values:

$1500=6t+0.9*t^{2}$ Solving for t we get 2 possible answers:

t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

8 0

3 years ago