Answer:
the magnitude of the angular magnification of the telescope. is 4
Explanation:
Calculate the magnitude of the angular magnification of the telescope.
Given that,
distance = 25cm
focal length from the objective f₀ = 20cm
focal length from eye piece f₁ = 5cm
The angular magnification of the telescope is

Magnification = 20 / 5
magnification = 4
Hence, the magnitude of the angular magnification of the telescope. is 4
Answer:
Explanation:
Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

Add them all together to get the x component of the resultant vector, V:

Do the same to find the y components of the part of this journey:

Add them together to get the y component of the resultant vector, V:

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.
We find the final magnitude:
and, rounding to 2 sig dig's as needed:
1.0 × 10² m; now for the direction:
58°
Answer:wdym insulation if u mean like covering than the ppls heads
Explanation:
Answer: The diameter of the circular path is 2.96m
Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.
Centripetal acceleration = 2.7m/s^2
Tangential speed = 2.0m/s
Radius = 2.0^2 / 2.7 = 4/2.7
= 1.48m
Diameter = radius*2
= 1.48*2 = 2.96m.