Infared = used by police
gamma = short wavelength
radio = largest wavelength
visible = only ones we can see
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
= 
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
= 
= 
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
learn more about acceleration here :
brainly.com/question/11560829
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Answer:
<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

- v is image distance
- u is object distance, u is 10 cm
- f is focal length, f is 5 cm

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>
• Let's derive this formula from the lens formula:

» Multiply throughout by fv

• But we know that, v/u is M

- v is image distance, v is 10 cm
- f is focal length, f is 5 cm
- M is magnification.

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>
- Image is magnified
- Image is erect or upright
- Image is inverted
- Image distance is identical to object distance.
Answer:
At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.
At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°
Explanation: