Witch type of radio station can broadcast a signal the gratis distance

Which statement best describes the circular flow model?

aleksley [76]

Can you please include the statement or the model?

3 0

2 years ago

A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H

Reika [66]

Answer:

The answer to your question is:

a) t1 = 2.99 s ≈ 3 s

b) vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ? time to reach the ground

vf = ? final speed

a) h = vot + (1/2)gt²

74 = 10t + (1/2)9.81t²

4.9t² + 10t -74 = 0 solve by using quadratic formula

t = (-b ± √ (b² -4ac) / 2a

t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

t = (-10 ± √ 1550.4 ) / 9.81

t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81

t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81

t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are

no negative times.

b) Formula vf = vo + gt

vf = 10 + (9.81)(3)

vf = 10 + 29.43

vf = 39.43 m/s

4 0

3 years ago

A capacitor is connected to an ac generator that has a frequency of 2.3 kHz and produces a rms voltage of 1.5 V. The rms current

miv72 [106K]

Answer:

Explanation:

Given that,

AC frequency of 2.3KHz

f=2.3×10³Hz

Vrms produce is

Vrms=1.5V

Current rms

Irms= 31mA

The capacitor is reconnected to a generator of frequency

f=4.8KHz =4800Hz

The current rms becomes

Irms= 85mA

Vrms=?

Solution

First genrator

The capacitive reactance is given as

Xc=Vrms/Irms

Xc=1.5/31×10^-3

Xc=48.39 ohms

Now, to know the capacitance of the capacitor

Xc=1/2πfC

Then,

C=1/2πfXc

So,

C=1/2×π×2300×48.39

C=1.43×10^-6C

C=1.43μF

Note: the capacitance of the capacitor did not change,

Now for generator two.

The reactance are given as

Xc=1/2πfC

Xc=1/2×π×4800×1.43×10^-6

Xc=23.19ohms

Then,

Vrms2=Irms2 ×Xc

Vrms2=85×10^-3×23.19ohms

Vrms2=1.97V

Vrms2=1.97Volts

7 0

3 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$