Answer:
<h3>On a velocity vs time graph the slope of the line represents the acceleration of the object. With a slope of zero, the object is moving at a constant velocity in the positive (+) direction during this five minute interval. ... Displacement and distance can both be determine on a velocity vs.</h3>
Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P = 
where
f = focal length
Thus
f = 
f =
= + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:

where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,



Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
Answer:
Explanation:
given,
cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.
vertical component = 9.28 cos∅
= 9.28 cos 27.27°
= 8.24 km
horizontal axis component = 9.28 sin ∅
= 9.28 sin 27.27°
= 4.5 km
distance of the final point from the origin
= 7.99 -(6.2-4.5)
= 6.29 km
displacement

d = 10.37 km
b) 
θ = 37.36°
Answer:
a) α=7.9x10^-4 rad
b) θ=1.12x10^-4 rad
c) The Earth and the Moon cannot be seen without a telescope.
Explanation:
In this exercise we will use the concepts of angular resolution, which depends on both the wavelength of the rays and the diameter of the eye or lens on the meter. Its unit of measure is the radian. The attached image shows the solution step by step.
Time = (distance) / (speed)
Time = (150 x 10⁹ m) / (3 x 10⁸ m/s) =
50 x 10¹ sec =
<em>500 sec</em> = 8 min 20 sec