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Vedmedyk [2.9K]

2 years ago

8

Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh

t-years. What is its luminosity?

1 answer:

iragen [17]2 years ago

6 0

For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

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What effects the amount of lift a plane gets?

vesna_86 [32]

Answer:

The airfoil shape and wing size will both affect the amount of lift. The ratio of the wing span to the wing area also affects the amount of lift generated by a wing. ... The lift then depends on the velocity of the air and how the object is inclined to the flow. Air: Lift depends on the mass of the flow.

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5 0

3 years ago

The drawing shows a model for the motion of the human forearm in throwing a dart. Because of the force M applied by the triceps

Serga [27]

Answer:

464.3 N

Explanation:

Given parameters are:

I = 0.065 $kg*m^2$

L = 0.025 m

R = 0.28 m

$v_0$ = 0 m/s

$v_f$ = 5 m/s

t = 0.1 s

$v_f=v_0+at=at$

Hence, $a=v_f/t$

We must connect two torque equations to find the answer.

$\tau=LM=I\alpha$

Where $\alpha =\frac{a}{R} =\frac{v_f}{Rt}$

Hence, $LM=I\frac{v_f}{Rt}$

Thus, $M = \frac{Iv_f}{LRt} = \frac{0.065*5}{0.025*0.28*0.1} =464.3$ N

5 0

4 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.6 multiply 1010 m (inside t

Doss [256]

We will apply the concepts related to energy conservation to develop this problem. In this way we will consider the distances and the given speed to calculate the final speed on the path from the sun. Assuming that the values exposed when saying 'multiply' is scientific notation we have the following,

$d_1 = 4.6*10^{10}m$

$v_i = 9.3*10^4m/s \rightarrow \text{Initial velocity comet}$

$d_2 = 6*10^{12}m$

The difference of the initial and final energy will be equivalent to the work done in the system, therefore

$E_f = E_i +W$

$K_f +U_f = K_i +U_i + 0$

$\frac{1}{2} mv_f^2+\frac{-GMm}{d_2} = \frac{1}{2} mv_i^2+\frac{-GMm}{d_1}$

Here,

m = Mass

$v_f$ = Final velocity

G = Gravitational Universal Constant

M = Mass of the Sun

m = Mass of the comet

$v_i$ = Initial Velocity

Rearranging to find the final velocity,

$v_f = \sqrt{v_i^2+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

Replacing with our values we have finally,

$v_f = \sqrt{(9.3*10^4)+2(6.7*10^{-11})(1.98*10^{30})(\frac{1}{6*10^{12}}-\frac{1}{4.6*10^{10}})}$

$v_f = 75653.9m/s$

Therefore the speed is 75653m/s

8 0

4 years ago

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

$Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta$

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

$E = (3)(230\ m/s)(0.61\ T)Sin90^o$

<u>E = 420.9 N/C</u>

3 0

3 years ago

Which can do work faster-a 700 watt gasoline engine or a 300 watt electric motor

iogann1982 [59]

"700 watts" means 700 joules of work per second.

"300 watts" means 300 joules of work per second.

If the labels on both machines are true, and both machines
are loaded to their full capacity, then the 700-watt engine
is doing work faster than the 300-watt one.

3 0

3 years ago