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gizmo_the_mogwai [7]
3 years ago
11

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial ExerciseA boat leaves a dock at 8:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 9:00 PM How many minutes after 8:00 PM were the two boats closest together
Physics
1 answer:
zubka84 [21]3 years ago
3 0

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

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Name any two liquid which are found in our human eyes​
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Aqueous humor and vitreous humor are the liquids present in the human eye.

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8 0
2 years ago
A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.
Serga [27]

Answer:

P = 4000 [Pa]

Explanation:

Pressure is defined as the relationship between Force and the area where the body rests.

The support area is equal to:

A=50*20=1000[cm^{2} ]

But we must convert from square centimeters to square meters.

1000[cm^{2}]*\frac{1^{2}m^{2}  }{100^{2}m^{2}  }=0.1[m^{2} ]

And the pressure is:

P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]

6 0
2 years ago
Which statement describes a controlled experiment?
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7 0
3 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
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