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7nadin3 [17]
3 years ago
13

A bird watcher spots a sparrow in a tree. The sparrow sits in a nest that is 10.5 feet above the bird watcher's eye level, at a

35° angle of elevation from the bird watcher. The bird watcher then notices a hawk in the same tree, 7.4 feet above the sparrow, at a certain angle of elevation. The bird watcher stands 15 feet from the base of the tree. What is the angle of elevation from the bird watcher to the hawk?
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Answer:

the angle is 50 degrees.

Explanation:

To solve this trigonometric problem we need ot remember that:

\alpha=arctg(\frac{O}{A})

where O is the opposite side of the triangle and A the adjacent

The hawk is 7.4 feet above the sparrow so:

O=10.5+7.4\\O=17.9ft

\alpha =arctg(\frac{17.9}{15})=50^o

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An object takes 2.8 years to orbit the Sun. What is its average distance (in AU) from the Sun?
Inessa05 [86]

answer look it up

Explanation:https://www.chegg.com/homework-help/questions-and-answers/takes-planet-28-years-orbit-sun-long-years-take-planet-go-way-around-sky-q29849397

5 0
2 years ago
A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. the horizontal compone
Strike441 [17]

When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.

Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

EMF = v.(B x L)

given that

B = 25\mu Tj - 50\mu Tk

v = 2 m/s j

L = 0.50 m (-i)

now by using the above formula we will have

EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)

EMF = 2(j) .(12.5\mu k + 25\mu j)

EMF = 50 \mu Volts

5 0
3 years ago
Read 2 more answers
As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh
MrMuchimi

Answer:37^{\circ} North of west

Explanation:

Given

40,000-ton luxury line traveling 20 knots towards west and

60,000 ton freighter traveling towards North with 10 knots

suppose v is the common velocity after collision

conserving momentum in west direction

20\times 40,000=(20,000+60,000)v\cos \theta

suppose the final velocity makes \theta angle with x axis

v\cos \theta =10----1

Conserving Momentum in North direction

10\times 60,000=80,000\times v\sin\theta

v\sin \theta =\frac{15}{2}----2

divide 1 and 2

\tan \theta =\frac{3}{4}

\theta =37^{\circ}

so search in the area 37^{\circ} North of west to find the ship

3 0
3 years ago
A 15.0 m long steel rod expands when its temperature rises from 34.0 degrees C to 50.0 degrees C. What is the change in the beam
azamat
0.00288................
8 0
3 years ago
Read 2 more answers
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
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