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3 years ago

5

Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi

nd conditions took 1h 24min. find the plane's air speed and the distance to the nearby town

2 answers:

insens350 [35]3 years ago

8 0

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

Oduvanchick [21]3 years ago

5 0

The answer to this question would be 60km per hour

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8 0

3 years ago

Read 2 more answers

Find the current flowing out of the battery.

klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0

3 years ago

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

The another equation of motion is $v^2=u^2+2as$ using this equation for second plane $40^2=0+2\times 2.67\times s$

s = 300 m

5 0

3 years ago

Which of these will be the correct relationship between work input and work output?

dsp73

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

8 0

2 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago