1) 14g
2) 24 fl oz
3)10 cups
4)12 and 1/2 cups
Given conditions:
height of object = 7.5cmdistance of object from mirror = 14 cmfocus length = -7 cmimage distance = ?
Using mirror formula:
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)
Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
Answer:
23.3 g/cm3
Explanation:
First of all, this should be in the math section. I just so happen to also be good at math. Cheers!
Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
½ k x² + m g x ’= ½ m v²
v² =
x² + g x
let's calculate
v² =
1.25² + 9.8 1.25
v² = 247.159 + 12.25 = 259.409
v = 16.11 m / s
Answer:
0.3405V
Explanation:
#Given a magnetic field of
, diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:

we can now calculate the induced emf,
:

Hence, the induced emf of the loop is 0.3405V