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Sindrei [870]
2 years ago
5

Two birds sit at the top of two different trees 17.8 feet away from one another. the distance between the first bird and a birdw

atcher on the ground is 37.4 feet. what is the angle measure, or angle of depression, between this bird and the birdwatcher? two trees have a bird on top of each. distance between birdwatcher and first bird is 37 point 4 feet. distance between each bird is 17 point 8 feet. angle of depression between birdwatcher and first bird is unknown. © 2011 jupiter images corporation 25.5° 64.5° 28.4° 61.6°
Physics
1 answer:
Yanka [14]2 years ago
7 0

The angle of depression between the bird and the birdwatcher is 61.32°

In the question, it is said that the distance between two birds is 17.8ft and the distance between the first bird and the birdwatcher is 37.4ft. So, when we draw an image in our mind then we can say that this whole system forms a right-angle triangle with a base of 17.8ft and a hypotenuse of 37.4ft. As we know that to calculate the angle of depression we will use trigonometric equations here i.e.,

cos x = Base/Hypotenuse

cos x = 17.8 / 37.4

cos x = 0.48

x = cos⁻¹(0.48)

x =  61.32°

Thus, the angle of depression is 61.32°.

Learn more about the angle of depression here:

brainly.com/question/26154210

#SPJ4

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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m
forsale [732]

Answer:

Acceleration due to gravity will be g=5.718m/sec^2

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing =\frac{145}{100}=1.45sec

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

T=2\pi \sqrt{\frac{l}{g}}

So 1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}

\sqrt{\frac{0.55}{g}}=0.230

Squaring both side

{\frac{0.3025}{g}}=0.0529

g=5.718m/sec^2

So acceleration due to gravity will be g=5.718m/sec^2

3 0
4 years ago
Which of the following is a conductor?<br><br> copper<br> water<br> aluminum<br> all of the above
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Answer:

D all of the above

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electricity moves easily through all of them and none of them prevent the flow of electricity

4 0
3 years ago
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3 years ago
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
3 years ago
how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?
SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

4 0
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