A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 10^6

Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th

Vladimir79 [104]

Answer:

$C = 1.77 \times 10^{-4} F$

Explanation:

As we know that in AC circuit we have

$V = i x_c$

here we have

V = 59 V

i = 5.05 A

so we will have

$x_c = \frac{59}{5.05}$

$x_c = 11.68 ohm$

also we know that

$x_c = \frac{1}{\omega C}$

here we will have

$11.68 = \frac{1}{(2\pi 77)C}$

$C = 1.77 \times 10^{-4} F$

7 0

2 years ago

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$

$\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})$

$\log_{10}(b_{1}/b_{2}) = 3.44$

Thus,

$(b_{1}/b_{2}) = 2754.22$

It means star A is 2754.22 time brighter than Star B.

3 0

3 years ago

Which of the following statements CANNOT be supported by Kepler's laws of planetary motion?

horsena [70]

Answer:

B) A planet's speed as it moves around the sun will not be the same in six months.

Explanation:

A planet's speed as it moves around the sun will not be the same in six months, is a statement that CANNOT be supported by Kepler's laws of planetary motion.

8 0

3 years ago

Read 2 more answers

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$