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2 years ago

How far away is the closest star?

Physics

2 answers:

sukhopar [10]2 years ago

8 0

Answer: Proxima Centauri is the closet star about 40,208,000,000,000 km away.

Explanation:

balu736 [363]2 years ago

4 0

Answer:

The closest star is about 25,300,000,000,000 miles (39,900,000,000,000 kilometers) away

Explanation:

You might be interested in

Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120

igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter.

As you have the circumference of B, you divide it by Pi to get the Diameter.

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter.

As' radius and Diameter will be 3x greater than B.

38.2 x 3 = 114.6

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 0

3 years ago

An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev

marishachu [46]

Answer:

Explanation:

Given that,

Force applied to pedal F = 50N

Angular velocity ω = 10rev/s

We know that, 1rev = 2πrad

Then, ω = 10rev/s = 10×2π rad/s

ω = 20π rad/s

Length of pedal r = 30cm = 0.3m

Power?

Power is given as

P = τ×ω

We need to find the torque τ

τ = r × F

Since r is perpendicular to F

Then, τ = 0.3 × 50

τ = 15 Nm

Then,

P = τ×ω

P = 15 × 20π

P = 942.48 Watts

power delivered to the bicycle by the athlete is 942.48 W

6 0

3 years ago

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

Anastasy [175]

Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now

6 0

3 years ago

A box of unknown mass is sliding with an initial speed vi = 4.70 m/s across a horizontal frictionless warehouse floor when it en

malfutka [58]

Here Change in Kinetic Energy = Work Done by Friction

Therefore, substituting the given values to the equation, we get

0.5 * m * (vFinal^2 - vInitial^2) = µ m g * d

Therefore

0.5*( 5.90^2 - Vfinal^2 ) = 0.100*9.8*2.10

Therefore

vfinal = 5.54 m/sec

8 0

3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago