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wariber [46]
4 years ago
15

An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),

the radar detects that the distance s is changing at a rate of 290 miles per hour. What is the speed of the airplane?

Physics
1 answer:
Novosadov [1.4K]4 years ago
5 0

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

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Answer:

Solution is given in the attachments

4 0
3 years ago
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
3 years ago
A motorcycle skids to a stop on a road.
Svetlanka [38]

Answer:

Friction of the road on the motorcycle in the opposite direction

Explanation:

Khanacademy

5 0
3 years ago
I'll mark brainliest
irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

                                  = 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

                                  = 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0
3 years ago
How can you use distance and displacement to describe an object's motion​
andrew11 [14]

Answer:

distance can describe the total distance moved and displacement shows how far something has moved from its starting position (in a straight line from point a to point b) the object doesn't have to move in a straight line, but that is how displacement is measured

8 0
3 years ago
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