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kkurt [141]
3 years ago
15

***FIRST CORRECT ANSWER GETS A BRAINLIEST***

Physics
2 answers:
dlinn [17]3 years ago
5 0
I think the answer is-D, it is halved. Hope I’ve helped ;)
melomori [17]3 years ago
3 0

Answer:

Explanation: It is halved

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You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb
mafiozo [28]

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

7 0
3 years ago
7. A golfer hits a golf ball a through a horizontal displacement of 80m. The time of flight for the ball was 1.8
Darya [45]

Answer:

44.4 m/s

Explanation:

d = 80 m, t = 1.8 s, find v

d = v*t

v = d/t = 80/1.8 = 44.4 m/s

6 0
2 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

3 0
3 years ago
Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t
MA_775_DIABLO [31]
First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200   Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x      Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x 
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time? 
Car A moves 20 meters every second:
20 x 40 = 800 meters
8 0
3 years ago
Since all stars begin their lives with the same basic composition, what characteristic most determines how they will differ?
Dafna1 [17]
4. Mass they are formed with
7 0
3 years ago
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