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inn [45]
2 years ago
6

A jet can travel the 6000 km distance between washington,

Physics
1 answer:
Andrei [34K]2 years ago
6 0
<span> d = r*t is the basic distance equation
 d = 6000 km
 t with the tail wind = 6 hr
 r with the tail wind = speed of the plane + wind speed = s + w
 t with the head wind = 7.5 hr
 r with the head wind = speed of the plane - wind speed = s-w
 (s+w)*6 = 6000
 (s-w)*7.5 = 6000
 s + w = 1000
 s - w = 800
</span><span> 2s = 1800
 s = 900 km/h
 s + w = 1000
 w = 100
 Check the anwer by calculating the return trip.
 (900-100) * 7.5 = 800 * 7.5
 800 * 7.5 = 6000 km
 Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>
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If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is

   √(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2  =  62.28 m/s .

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6 0
3 years ago
A sound wave was determined to have a frequency of 0.3 Hz, speed of 150 cm/s, and amplitude of 2 cm. Find its wavelength.
fredd [130]

Answer:

5 m

Explanation:

From the question,

v = λf....................... Equation 1

Where v = speed of the sound wave, λ = wavelength of the sound wave, f = frequency of the sound wave.

make λ the subject of the equation

λ = v/f..................... Equation 2

Given: v = 150 cm/s = 1.5 m/s, f = 0.3 hz.

Substitute these values into equation 2

λ = 1.5/0.3

λ = 5 m.

7 0
2 years ago
A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li
saveliy_v [14]

Answer:

29.2 ft/s

Explanation:

The distance of the light's projection on the wall

y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,

Angular speed = w = (2π/3)

w = (θ/t)

θ = wt = (2πt/3)

(dθ/dt) = (2π/3)

y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

(dy/dt) = (26π/3) sec² (15°)

(dy/dt) = 29.2 ft/s

5 0
3 years ago
See attachment for question
zloy xaker [14]

The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.

The ratio of the calculated and accepted value of the Mercury's mass is 0.95.

<h3>What is mass?</h3>

Mass is the amount of matter present in the object.

The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.

Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.

The acceleration due to gravity is related with mass as

g = GM/R²

Substitute the values, we have

3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³

M = 2.2016 x 10¹³ /  6.67 x 10⁻¹¹

M = 0.3152 x 10²⁴ kg

Thus, the mercury's mass is  0.3152 x 10²⁴ kg.

(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg

Ratio of the value of mass calculated and accepted is

Mcalc/M accep =  0.3152 x 10²⁴ kg / 3.301 x 10²³ kg

                          = 0.95

Thus, the ratio is 0.95

Learn more about mass.

brainly.com/question/19694949

#SPJ1

8 0
1 year ago
if the separation distance between the moon and the planet is increased by a factor of 4 then the force gravitational is
erastova [34]

Answer:

Fg = (G * m1 * m2) / r^2

The gravitational force is Fg/16

Explanation:

Fg = (G * m1 * m2) / (4r)^2

Fg = (G * m1 * m2) / 16r

Therefore,

Fg / 16

8 0
3 years ago
Read 2 more answers
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