Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
Answer:
no not always sometimes they react at all so false I hope I helped :)
From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
The given weight of Elliot is 600 N
From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.
P.E = mgh
where mg = Weight = 600
To find the height Elliot will reach, substitute all necessary parameters into the equation above.
250 = 600h
Make h the subject of the formula
h = 250/600
h = 0.4167 meters
Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
Learn more about energy here: brainly.com/question/24116470
Answer:
304 meters downstream
Explanation:
The given parameters are;
The speed of the swimmer = 2.00 m/s
The width of the river = 73.0 m
The speed of the river = 8.00 m/s
Therefore;
The direction of the swimmer's resultant velocity = tan⁻¹(8/2) ≈ 75.96° downstream
The distance downstream the swimmer will reach the opposite shore = 4 × 73 = 304 m downstream
The distance downstream the swimmer will reach the opposite shore = 304 m downstream
Here's a quick way to find out. Pick up your glasses, bifocals work best, and find the focal length with a flashlight against a book. If I remember right, the object should be magnified and upside down. So, A.